Free math worksheets

- A series converges when its sequence of partial sums approaches a finite limit. - Separate the constant part of \(S_n\) from the fraction that depends on \(n\). - Determine what happens to that fraction as \(n\to\infty\), then interpret the resulting partial-sum limit.

1. The infinite series converges exactly when the sequence \(S_n\) has a finite limit. 2. The correction term tends to \(0\). 3. Thus \(\lim_{n\to\infty}S_n=7\), so the series converges to \(7\).

The series converges to \(7\).

- Examine how \(\ln(n+1)\) behaves as \(n\) increases without bound. - Convergence of the series requires a finite limit for the partial sums. - Decide whether an unbounded partial-sum sequence can satisfy that requirement.

1. The partial sums increase without bound. 2. Therefore the sequence of partial sums has no finite limit. 3. The series diverges.

The series diverges.

- Treat \(2^{-n}\) as \(\left(\frac12\right)^n\) and recall its long-term behavior. - Find the limit of the given partial-sum formula rather than the limit of the individual series terms. - A finite partial-sum limit is the value of the infinite series.

1. The infinite series converges exactly when the sequence \(S_n\) has a finite limit. 2. The exponential correction tends to \(0\). 3. Thus \(\lim_{n\to\infty}S_n=5\), so the series converges to \(5\).

The series converges to \(5\).

- Determine the limit of the partial sums directly from the denominator’s growth. - Remember that a series may converge to \(0\); the sum need not be positive or nonzero. - State the series conclusion from the finite limit of \(S_n\).

1. The infinite series converges exactly when the sequence \(S_n\) has a finite limit. 2. The partial sums approach \(0\). 3. Thus \(\lim_{n\to\infty}S_n=0\), so the series converges to \(0\).

The series converges to \(0\).

- The factor \((-1)^n\) changes sign, but its absolute value stays equal to \(1\). - Bound the magnitude of the oscillating correction by a simpler expression that approaches \(0\). - Use the resulting limit of \(S_n\) to decide whether the series has a finite sum.

1. The infinite series converges exactly when the sequence \(S_n\) has a finite limit. 2. The oscillation has magnitude \(\frac{1}{n}\to0\). 3. Thus \(\lim_{n\to\infty}S_n=4\), so the series converges to \(4\).

The series converges to \(4\).

- Evaluate the sine term for several consecutive integer values of \(n\). - Look for a repeating pattern in the partial sums rather than averaging their values. - A convergent series requires the entire partial-sum sequence to approach one number.

1. The partial sums cycle through \(3,2,1,2,\ldots\). 2. Therefore the sequence of partial sums has no finite limit. 3. The series diverges.

The series diverges.

- Divide the numerator and denominator by the highest power of \(n\) present. - Identify which terms remain relevant as \(n\to\infty\). - Interpret the resulting finite limit as the sum of the series.

1. The infinite series converges exactly when the sequence \(S_n\) has a finite limit. 2. The rational expression approaches the ratio of leading coefficients. 3. Thus \(\lim_{n\to\infty}S_n=4\), so the series converges to \(4\).

The series converges to \(4\).

- For integer \(n\), the cosine factor is bounded even though it oscillates. - Compare the absolute value of the correction term with \(\frac1{\sqrt n}\). - Use that bound to determine the limit of the partial sums.

1. The infinite series converges exactly when the sequence \(S_n\) has a finite limit. 2. The oscillating correction has magnitude \(\frac{1}{\sqrt n}\to0\). 3. Thus \(\lim_{n\to\infty}S_n=6\), so the series converges to \(6\).

The series converges to \(6\).

- Evaluate \(S_n\) separately for even and odd values of \(n\). - Check whether those two subsequences approach the same number. - Persistent oscillation between distinct values prevents convergence of the series.

1. The partial sums alternate between \(45\) and \(55\). 2. Therefore the sequence of partial sums has no finite limit. 3. The series diverges.

The series diverges.

- One parity subsequence is constant; identify its limit immediately. - Determine what happens to the reciprocal correction in the other parity subsequence. - Compare the two subsequential limits before concluding anything about the series.

1. The even partial sums are constantly \(1\). 2. The odd partial sums satisfy \(1+\frac{1}{m}\to1\). 3. Thus \(S_n\to1\), so the series converges to \(1\).

The series converges to \(1\).

- Generate the next few partial sums from the recurrence. - Check whether those values settle toward one number or enter a repeating cycle. - A nonconstant cycle in the partial sums prevents the series from converging.

1. \(S_1=2\), \(S_2=4\), and \(S_3=2\). 2. The partial sums repeat the cycle \(2,4,2,4,\ldots\). 3. The sequence has no limit, so the series diverges.

The series diverges.

- For a rational expression with equal-degree numerator and denominator, the limit depends on the leading coefficients. - Express the limiting partial sum in terms of \(c\). - Set that limit equal to the required series sum and solve the resulting equation.

1. \(\lim_{n\to\infty}S_n=\frac{c}{2}\). 2. Set \(\frac{c}{2}=3\). 3. Thus \(c=6\).

\(c=6\).

- First find the limiting partial sum in terms of \(a\) by comparing leading coefficients. - Translate “converges to a negative number” into an inequality for that limit. - Solve the inequality, noting that the denominator coefficient is positive.

1. \(\lim_{n\to\infty}S_n=\frac{a}{3}\). 2. The limit is negative exactly when \(\frac{a}{3}<0\). 3. Thus \(a<0\).

\(a<0\).

- Boundedness and convergence are different properties of a sequence. - List the values of \(S_n\) for several even and odd indices. - Decide whether the partial sums approach one number or continue oscillating.

1. The partial sums alternate between \(1\) and \(-1\). 2. Boundedness alone does not imply that a sequence has a limit. 3. The series diverges.

The claim is false; the series diverges.

- Compare each new partial sum with the corresponding original partial sum. - Changing only the first term shifts every partial sum by the same constant. - Apply that same shift to the original limiting sum.

1. Every new partial sum is \(7\) less than the corresponding original partial sum. 2. The new limit is \(12-7=5\). 3. The new series converges to \(5\).

The new series converges to \(5\).

- Use the given limit of \(S_n\) together with the limit of \(\frac1n\). - Apply limit laws to the sum and difference defining \(T_n\). - The limit of \(T_n\), when finite, is the sum of series \(B\).

1. \(S_n\to-4\) and \(1/n\to0\). 2. Therefore \(T_n\to-4+9-0=5\). 3. Series \(B\) converges to \(5\).

Series \(B\) converges to \(5\).

- Compare the distance \(|S_n-5|\) with a sequence whose limit is known. - The sum of a series is the limit of its partial sums. - Solve the error inequality using the given upper bound, then choose the first integer strictly above the cutoff.

1. Because \(0\le |S_n-5|\le \frac{4}{n^2}\) and \(\frac{4}{n^2}\to0\), the Squeeze Theorem gives \(|S_n-5|\to0\). 2. Therefore \(S_n\to5\), so the series converges to \(5\). 3. To guarantee \(|S_n-5|<0.01\), it is enough to require \(\frac{4}{n^2}<0.01\). 4. This inequality is equivalent to \(n^2>400\), so the least positive integer that works for every later index is \(N=21\).

The series converges to \(5\), and the least integer guaranteed by the bound is \(N=21\).

- Separate each new term into its two original contributions. - Compare the shifted copy of the original series with the full series. - Account for the one term that disappears after the shift.

1. Split the new series as \(\sum_{n=1}^{\infty}b_n=\sum_{n=1}^{\infty}a_n+\sum_{n=1}^{\infty}a_{n+1}\). 2. The shifted series omits only \(a_1\), so \(\sum_{n=1}^{\infty}a_{n+1}=8-3=5\). 3. Both component series converge, and their sums add to \(8+5=13\).

The series \(\sum_{n=1}^{\infty}b_n\) converges to \(13\).

- Compare the new partial sums with the original partial sums. - Determine how many consecutive new partial sums share the same value. - Inserting zero terms changes positions but not accumulated totals.

1. Let \(S_n=\sum_{k=1}^{n}a_k\) and \(T_m=\sum_{k=1}^{m}b_k\). 2. For each \(n\), the three partial sums \(T_{3n-2}\), \(T_{3n-1}\), and \(T_{3n}\) all equal \(S_n\). 3. Since \(S_n\to L\), every sufficiently late partial sum of the new series equals a sufficiently late partial sum of the original series. 4. Therefore \(T_m\to L\).

The zero-inserted series converges to \(L\).

- How do nonnegative terms affect consecutive partial sums? - Combine the direction of movement with the supplied bound. - Decide whether the information determines a unique limiting value.

1. Since \(a_n\ge0\), \(S_{n+1}=S_n+a_{n+1}\ge S_n\), so the partial-sum sequence is nondecreasing. 2. The partial sums are bounded above by \(12\). 3. Every nondecreasing sequence that is bounded above converges. 4. Therefore the series converges to some sum \(L\) satisfying \(0\le L\le12\). The exact value cannot be determined from the given information.

The series converges to some \(L\) with \(0\le L\le12\); the exact sum is not determined.

- Track the total effect of each changed term on sufficiently long partial sums. - After the last changed position, how are the two partial-sum sequences related? - A fixed vertical shift changes a limit by the same amount.

1. The two changes alter the total of every partial sum with index at least \(10\) by \(7-2=5\). 2. If \(S_n\) and \(T_n\) are the original and new partial sums, then \(T_n=S_n+5\) for every \(n\ge10\). 3. Since \(S_n\to L\), it follows that \(T_n\to L+5\). 4. Therefore the modified series converges to \(L+5\).

The modified series converges to \(L+5\).

- Compare the unknown partial sum directly with the known limiting value. - Insert the other partial sum between those two quantities. - Use the given error bound together with the known convergence.

1. By the triangle inequality, \(|T_n-5|\le|T_n-S_n|+|S_n-5|\). 2. The first term is at most \(\frac{2}{n}\to0\), and the second term approaches \(0\) because \(S_n\to5\). 3. Therefore \(|T_n-5|\to0\), so \(T_n\to5\).

The second series converges to \(5\).

- Remove the subtraction between two nearly equal radical expressions. - Scale the resulting expression by the dominant power of the index. - Evaluate the limit of the simplified partial sum.

1. Rationalize the difference: \(S_n=n\cdot\frac{1}{\sqrt{n^2+1}+n}\). 2. Divide numerator and denominator by \(n\) to obtain \(S_n=\frac{1}{\sqrt{1+1/n^2}+1}\). 3. Taking the limit gives \(S_n\to\frac{1}{2}\).

The series converges to \(\frac{1}{2}\).

- Compare the sets of terms contained in sufficiently long partial sums. - Order does not affect a finite sum. - Determine from which index onward the two partial-sum sequences agree exactly.

1. A rearrangement of the first \(20\) terms has the same finite total \(a_1+\cdots+a_{20}\). 2. For every \(n\ge20\), the new \(n\)th partial sum contains exactly the same first \(n\) terms as the original \(n\)th partial sum, only with the first \(20\) in a different order. 3. Therefore the two partial sums are equal for all \(n\ge20\). 4. Since the original partial sums approach \(L\), the new partial sums also approach \(L\).

The rearranged series converges to the same sum \(L\).

- Factor the denominator. - Use partial fractions to write the term as a difference. - Expand several terms and identify the cancellations. - Take the limit of the remaining endpoint terms.

1. Factor the denominator and decompose the term: \(\frac{2}{k(k+2)}=\frac1k-\frac{1}{k+2}\). 2. The \(n\)th partial sum is \(S_n=\sum_{k=1}^{n}\left(\frac1k-\frac{1}{k+2}\right)\). Writing out the terms shows that the series telescopes: \(S_n=1+\frac12-\frac{1}{n+1}-\frac{1}{n+2}\). 3. Take the limit: \(\lim_{n\to\infty}S_n=1+\frac12=\frac32\).

\(\frac32\).

- Factor the difference of squares in the denominator. - Decompose the term into a difference of fractions. - Write the first few terms to expose the telescoping pattern. - Evaluate the remaining expression as \(n\to\infty\).

1. Factor the denominator: \(4k^2-1=(2k-1)(2k+1)\). 2. Use partial fractions: \(\frac{1}{4k^2-1}=\frac12\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)\). 3. The \(n\)th partial sum telescopes: \(S_n=\frac12\left[\left(1-\frac13\right)+\left(\frac13-\frac15\right)+\cdots+\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)\right]\) \(=\frac12\left(1-\frac{1}{2n+1}\right)\). 4. Therefore, \(\lim_{n\to\infty}S_n=\frac12\).

\(\frac12\).

- Analyze the partial sums separately when \(n\) is even and when \(n\) is odd. - Find the limiting value of each of those two subsequences. - If the subsequences approach different numbers, the full partial-sum sequence cannot converge.

1. Even partial sums approach \(1\) while odd partial sums approach \(-1\). 2. Therefore the sequence of partial sums has no finite limit. 3. The series diverges.

The series diverges.

- Compute the first term from \(a_1=S_1\); the difference formula is only needed after that. - For \(n\ge2\), use \(a_n=S_n-S_{n-1}\) and combine the two rational expressions over a common denominator. - Determine the series sum by taking the limit of the given partial-sum formula.

1. a) \(a_1=S_1=\frac{1}{3}\). 2. a) For \(n\ge2\), \(a_n=S_n-S_{n-1}=\frac{2}{\left(n + 1\right) \left(n + 2\right)}\). 3. b) \(\lim_{n\to\infty}S_n=1\), so the series converges to \(1\).

a) \(a_1=\frac{1}{3}\), and \(a_n=\frac{2}{\left(n + 1\right) \left(n + 2\right)}\) for \(n\ge2\). b) The sum is \(1\).

- Start with \(a_1=S_1\), using the partial-sum formula at \(n=1\). - For later terms, subtract \(S_{n-1}\) from \(S_n\) carefully and simplify the numerator after finding a common denominator. - For the total sum, compare the leading terms in the numerator and denominator of \(S_n\).

1. a) \(a_1=S_1=\frac{4}{7}\). 2. a) For \(n\ge2\), \(a_n=S_n-S_{n-1}=\frac{13}{\left(2 n + 3\right) \left(2 n + 5\right)}\). 3. b) \(\lim_{n\to\infty}S_n=\frac{3}{2}\), so the series converges to \(\frac{3}{2}\).

a) \(a_1=\frac{4}{7}\), and \(a_n=\frac{13}{\left(2 n + 3\right) \left(2 n + 5\right)}\) for \(n\ge2\). b) The sum is \(\frac{3}{2}\).

- Obtain \(a_1\) directly from the first partial sum. - In \(S_n-S_{n-1}\), the constant parts cancel, leaving a difference of two reciprocals. - The sum of the series is the limit of the constant term plus the vanishing reciprocal term.

1. a) \(a_1=S_1=5\). 2. a) For \(n\ge2\), \(a_n=S_n-S_{n-1}=- \frac{3}{n \left(n - 1\right)}\). 3. b) \(\lim_{n\to\infty}S_n=2\), so the series converges to \(2\).

a) \(a_1=5\), and \(a_n=- \frac{3}{n \left(n - 1\right)}\) for \(n\ge2\). b) The sum is \(2\).

- Evaluate \(S_1\) before applying the consecutive-partial-sum formula. - For \(n\ge2\), write \(S_{n-1}\) by replacing every \(n\) in the formula with \(n-1\), then subtract. - Use the ratio of leading coefficients to find the limiting partial sum.

1. a) \(a_1=S_1=0\). 2. a) For \(n\ge2\), \(a_n=S_n-S_{n-1}=\frac{2}{n \left(n + 1\right)}\). 3. b) \(\lim_{n\to\infty}S_n=1\), so the series converges to \(1\).

a) \(a_1=0\), and \(a_n=\frac{2}{n \left(n + 1\right)}\) for \(n\ge2\). b) The sum is \(1\).

- Find \(a_1\) from \(S_1\) rather than using a difference involving \(S_0\). - When computing \(S_n-S_{n-1}\), track the minus signs in the reciprocal terms. - The reciprocal correction vanishes in the limit, leaving the series sum.

1. a) \(a_1=S_1=6\). 2. a) For \(n\ge2\), \(a_n=S_n-S_{n-1}=\frac{4}{n \left(n - 1\right)}\). 3. b) \(\lim_{n\to\infty}S_n=10\), so the series converges to \(10\).

a) \(a_1=6\), and \(a_n=\frac{4}{n \left(n - 1\right)}\) for \(n\ge2\). b) The sum is \(10\).

- Treat the even-indexed and odd-indexed partial sums as two subsequences. - Find the limit of each formula as \(m\to\infty\). - The full sequence converges when these two subsequences approach the same value.

1. \(S_{2m}\to2\). 2. \(S_{2m-1}\to2\). 3. Both parity subsequences have the same limit, so \(S_n\to2\). 4. The series converges to \(2\).

The series converges to \(2\).

- Compute the limiting behavior of the even partial sums and odd partial sums separately. - Ask whether both subsequences approach the same candidate limit. - Two different subsequential limits rule out convergence of the entire partial-sum sequence.

1. \(S_{2m}\to2\). 2. \(S_{2m-1}\to3\). 3. The two subsequences have different limits, so \(S_n\) has no limit. 4. The series diverges.

The series diverges.

- Divide every term in the numerator and denominator by \(n^2\). - Treat \(k\) as a fixed real number when taking the limit. - Decide whether any term containing \(k\) survives as \(n\to\infty\).

1. Divide by \(n^2\): \(S_n=\frac{1+k/n}{1+1/n^2}\). 2. For every fixed real \(k\), \(k/n\to0\) and \(1/n^2\to0\). 3. Therefore \(S_n\to1\) for every real \(k\).

The series converges to \(1\) for every real \(k\).

- Analyze the three stated cases separately; the behavior of \(n^p\) changes with the sign of \(p\). - For negative \(p\), rewrite \(n^p\) as a reciprocal power. - Substitute the limiting behavior of \(n^p\) into the partial-sum expression in each case.

1. For \(p>0\), \(n^p\to\infty\), so \(S_n\to1\). 2. For \(p=0\), \(S_n=\frac12\). 3. For \(p<0\), \(n^p\to0\), so \(S_n\to0\).

For \(p>0\), the sum is \(1\). For \(p=0\), the sum is \(\frac12\). For \(p<0\), the sum is \(0\).

- Distinguish numerical evidence from a proof about all sufficiently large indices. - Ask whether another sequence could match every displayed value and then behave differently afterward. - A convergence proof needs a general formula, bound, or argument—not only finitely many data points.

1. The values are consistent with convergence to \(2\), but only finitely many partial sums are displayed. 2. A sequence can match all four entries and behave differently afterward. 3. Therefore the table does not prove convergence.

No. The table suggests a sum of \(2\), but it does not prove it.

- Rewrite \(|S_n-8|\) as a simple expression in \(n\). - Solve the strict inequality for \(n\). - Which least integer works for that value and every larger index?

1. \(|S_n-8|=\frac{5}{n+1}\). 2. \(\frac{5}{n+1}<0.01\) is equivalent to \(n+1>500\). 3. The least integer is \(N=500\).

\(N=500\).

- Taking the absolute value removes the alternating sign. - Compare the reciprocal error with the tolerance, keeping the inequality strict. - After solving for the index, choose the least integer that also makes every later error smaller.

1. \(|S_n-3|=\frac{1}{2n+1}\). 2. \(\frac{1}{2n+1}<0.005=\frac1{200}\) requires \(2n+1>200\). 3. The least integer is \(N=100\).

\(N=100\).

- What happens to the upper bound as \(n\) increases? - Which theorem applies when a nonnegative expression is bounded above by a sequence tending to \(0\)? - How does \(|S_n-7|\to0\) identify the series sum?

1. \(\frac{3}{\sqrt n}\to0\). 2. The inequality gives \(0\le|S_n-7|\le\frac{3}{\sqrt n}\). 3. By the Squeeze Theorem, \(|S_n-7|\to0\), so \(S_n\to7\). 4. The series converges to \(7\).

The series converges to \(7\).

- Combine neighboring terms and examine what remains from each pair. - Compare the partial sums that end after a complete pair with those that stop one term earlier. - A full sequence converges when its even and odd subsequences approach the same value.

1. Each two-term block has sum \(a_{2k-1}+a_{2k}=\frac{1}{2^k}\). 2. Thus \(S_{2m}=\sum_{k=1}^{m}\frac{1}{2^k}=1-\frac{1}{2^m}\), so \(S_{2m}\to1\). 3. Since \(S_{2m-1}=S_{2m}-a_{2m}=1+\frac{1}{m^2}-\frac{2}{2^m}\), the odd partial sums also approach \(1\). 4. Both parity subsequences of partial sums have the same limit, so the series converges to \(1\).

The series converges, and its sum is \(1\).

- Turn the sum of logarithms into the logarithm of a product. - Write several factors of the finite product and identify the cancellations. - Take the limit only after simplifying the complete partial sum.

1. The \(N\)th partial sum is \(S_N=\ln\left(\prod_{n=1}^{N}\frac{(n+1)^2}{n(n+2)}\right)\). 2. Cancelling common factors in the product gives \(\prod_{n=1}^{N}\frac{(n+1)^2}{n(n+2)}=\frac{2(N+1)}{N+2}\). 3. Therefore \(S_N=\ln\left(\frac{2(N+1)}{N+2}\right)\). 4. Since \(\frac{2(N+1)}{N+2}\to2\), the series converges to \(\ln 2\).

The series converges to \(\ln 2\).

- How can the possible index positions be organized into a small number of repeating cases? - What does the long-term behavior look like in each case? - Do those cases together cover every partial sum?

1. As \(m\to\infty\), \(S_{3m-2}\to4\) because \(\frac{1}{m}\to0\). 2. Also, \(S_{3m-1}\to4\) because \(\frac{1}{m^2}\to0\), and \(S_{3m}\to4\) because \(\frac{(-1)^m}{m}\to0\). 3. Every partial sum belongs to exactly one of these three subsequences, and all three subsequences approach \(4\). 4. Hence \(S_n\to4\), so the series converges and has sum \(4\).

The series converges to \(4\).

- What happens to the adjacent intervals when several terms are added? - Can the finite sum be combined before any endpoint behavior is examined? - Which part of the resulting endpoint expression controls the limit?

1. Adjacent integration intervals combine, so \(S_N=\sum_{n=1}^{N}a_n=\int_{0}^{N}x e^{-x}\,dx\). 2. An antiderivative is \(-(x+1)e^{-x}\), giving \(S_N=1-(N+1)e^{-N}\). 3. Because \((N+1)e^{-N}\to0\), the partial sums approach \(1\).

The series converges to \(1\).

- Organize the terms by whether their positions are odd or even. - Examine partial sums that stop after each type of position. - Determine whether all resulting subsequences approach the same value.

1. Let \(O_m=\sum_{n=1}^{m}a_{2n-1}\) and \(E_m=\sum_{n=1}^{m}a_{2n}\), with \(E_0=0\). Then \(O_m\to A\) and \(E_m\to B\). 2. The even partial sums of the original series satisfy \(S_{2m}=O_m+E_m\to A+B\). 3. The odd partial sums satisfy \(S_{2m-1}=O_m+E_{m-1}\to A+B\). 4. Both parity subsequences approach the same limit, so \(S_n\to A+B\).

The original series converges to \(A+B\).

- Bound the greatest-integer expression using the number immediately below it. - Divide the bounds by the same positive denominator. - Compare the limiting values of the lower and upper bounds.

1. The floor inequality gives \(\sqrt n-1<\lfloor\sqrt n\rfloor\le\sqrt n\). 2. Dividing by \(\sqrt n\) gives \(1-\frac{1}{\sqrt n}<S_n\le1\). 3. Both bounds approach \(1\), so \(S_n\to1\) by the squeeze theorem. 4. Therefore the series converges to \(1\).

The series converges to \(1\).

- What fixed interval is represented by the finite sum? - Which sample points and widths are encoded in the expression? - Remember that the relevant limit belongs to the partial sums themselves.

1. Write \(S_n=\sum_{k=1}^{n}\frac{1}{1+(k/n)^2}\cdot\frac{1}{n}\). 2. This is the right-endpoint Riemann sum for \(f(x)=\frac{1}{1+x^2}\) on \([0,1]\). 3. Hence \(\lim_{n\to\infty}S_n=\int_0^1\frac{1}{1+x^2}\,dx=\arctan(1)-\arctan(0)=\frac{\pi}{4}\). 4. The series therefore converges to \(\frac{\pi}{4}\).

The series converges to \(\frac{\pi}{4}\).

- Place each odd-indexed partial sum between neighboring even-indexed partial sums. - Determine what happens to both neighboring bounds as the index grows. - Combine the conclusions for the two parity subsequences.

1. For each \(m\), monotonicity gives \(S_{2m}\le S_{2m+1}\le S_{2m+2}\). 2. The lower bound \(S_{2m}\) approaches \(6\), and the upper bound \(S_{2m+2}\) is a shifted even-indexed subsequence that also approaches \(6\). 3. Therefore \(S_{2m+1}\to6\) by the squeeze theorem. 4. Both the even and odd partial sums approach \(6\), so \(S_n\to6\) and the series converges to \(6\).

The series converges to \(6\).

- Recall what convergence of the original series forces its individual terms to do. - Expand a finite sum of consecutive differences. - Identify the two endpoint terms that remain.

1. Since \(\sum a_n\) converges, its terms satisfy \(a_n\to0\). 2. The \(N\)th partial sum of the new series is \(T_N=\sum_{n=1}^{N}(a_n-a_{n+1})=a_1-a_{N+1}\). 3. Taking the limit gives \(T_N\to a_1\). 4. Thus \(\sum b_n\) converges to \(a_1\).

The series \(\sum_{n=1}^{\infty}b_n\) converges to \(a_1\).

- Combine the finite number of integrals before evaluating them. - Simplify the finite power sum after multiplying by the remaining factor. - Evaluate the resulting elementary integral and then take its limit.

1. For a finite partial sum, \(S_N=\int_0^1(1-x)\sum_{n=1}^{N}x^{n-1}\,dx\). 2. Since \((1-x)\sum_{n=1}^{N}x^{n-1}=1-x^N\), \(S_N=\int_0^1(1-x^N)\,dx\). 3. Therefore \(S_N=1-\frac{1}{N+1}=\frac{N}{N+1}\to1\).

The series converges to \(1\).

- Apply the given bound repeatedly to earlier partial sums. - Look for a multiplicative pattern in the distance from the proposed limit. - Determine what happens to that bound after many repetitions.

1. Repeatedly apply the inequality to obtain \(|S_n-2|\le\left(\frac12\right)^{n-1}|S_1-2|\). 2. The right-hand side approaches \(0\) as \(n\to\infty\). 3. Therefore \(|S_n-2|\to0\), which means \(S_n\to2\). 4. By the definition of an infinite-series sum, the series converges to \(2\).

The series converges to \(2\).

- Examine the partial sums or term limits of the two component series separately. - Simplify the term obtained after adding the two sequences. - Rewrite the remaining rational term to reveal cancellation in its partial sums.

1. Since \(a_n=1\), the partial sums of \(\sum a_n\) equal \(N\) and are unbounded, so the series diverges. 2. Since \(b_n\to-1\ne0\), \(\sum b_n\) diverges. 3. The combined term is \(a_n+b_n=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\). 4. Its \(N\)th partial sum is \(1-\frac{1}{N+1}\to1\), so \(\sum(a_n+b_n)\) converges to \(1\).

Both \(\sum a_n\) and \(\sum b_n\) diverge, while \(\sum(a_n+b_n)\) converges to \(1\).

- Rewrite the average using the change from one partial sum to the next. - Isolate the original partial sum. - Combine the two given limiting behaviors.

1. Since \(S_{n+1}=S_n+a_{n+1}\), \(T_n=S_n+\frac{a_{n+1}}{2}\). 2. Rearranging gives \(S_n=T_n-\frac{a_{n+1}}{2}\). 3. The first term on the right approaches \(L\), and the second approaches \(0\). 4. Hence \(S_n\to L\), so the series converges to \(L\).

The series converges to \(L\).

- Simplify the finite rational sum by rewriting each summand as a difference. - Bound the oscillating correction by a nonoscillating quantity. - Combine the limits of all remaining pieces.

1. Since \(\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}\), the finite sum equals \(1-\frac{1}{n+1}\). 2. Thus \(S_n=1-\frac{1}{n+1}+\frac{\cos n}{n}\). 3. The second correction approaches \(0\), and \(\left|\frac{\cos n}{n}\right|\le\frac{1}{n}\to0\). 4. Therefore \(S_n\to1\), so the series converges to \(1\).

The series converges to \(1\).

- Separate the new series into the original terms and the added corrections. - Reindex the corrections by the square root of their locations. - Find the total contribution of the correction series.

1. The new series is the original series plus one added term \(\frac{1}{2^m}\) at each square index \(m^2\). 2. The total added contribution is \(\sum_{m=1}^{\infty}\frac{1}{2^m}=1\). 3. Both the original series and the added series converge, so the new series converges to \(4+1=5\).

The series \(\sum_{n=1}^{\infty}c_n\) converges to \(5\).

- Express the index as the reciprocal of the small angle. - Introduce a second power of the small angle to match a standard trigonometric limit. - Track the remaining factor after applying that limit.

1. Rewrite \(S_n=\frac{1-\cos(1/n)}{1/n}\). 2. Multiply and divide by \(1/n\): \(S_n=\left(\frac{1-\cos(1/n)}{(1/n)^2}\right)\frac{1}{n}\). 3. The first factor approaches \(\frac12\), while the second approaches \(0\). 4. Hence \(S_n\to0\), so the series converges to \(0\).

The series converges to \(0\).

- Recover a sequence term by subtracting consecutive partial sums. - Decompose the rational expression into partial fractions. - Write several terms of the sum and look for cancellation. - Evaluate the remaining term as \(n\to\infty\).

1. For \(n>1\), \(a_n=S_n-S_{n-1}\) \(=(2n^2+n)-[2(n-1)^2+(n-1)]=4n-1\). For \(n=1\), \(a_1=S_1=3\), which also agrees with the formula. 2. Since \(a_n=4n-1\) and \(a_{n+1}=4n+3\), \(b_n=\frac{1}{(4n-1)(4n+3)}=\frac14\left(\frac{1}{4n-1}-\frac{1}{4n+3}\right)\). 3. The partial sums telescope: \(T_n=\frac14\left[\left(\frac13-\frac17\right)+\left(\frac17-\frac1{11}\right)+\cdots+\left(\frac{1}{4n-1}-\frac{1}{4n+3}\right)\right]\) \(=\frac14\left(\frac13-\frac{1}{4n+3}\right)\). 4. Since \(\frac{1}{4n+3}\to0\), \(\lim_{n\to\infty}T_n=\frac14\cdot\frac13=\frac1{12}\).

1) \(a_n=4n-1\). 2) \(T_n=\frac14\left(\frac13-\frac{1}{4n+3}\right)\), and \(\lim_{n\to\infty}T_n=\frac1{12}\).

- If the partial sums approach a limit \(L\), what equation must \(L\) satisfy? - What recurrence results after subtracting the candidate limit from both sides? - Does the resulting geometric factor tend to \(0\)?

1. A limiting value \(L\) must satisfy \(L=\frac25L+3\), so \(L=5\). 2. Subtracting \(5\) gives \(S_n-5=\frac25(S_{n-1}-5)\). 3. Thus \(S_n-5=-4\left(\frac25\right)^{n-1}\to0\). 4. The series converges to \(5\).

The series converges to \(5\).

- Estimate the change in the partial-sum function over one unit of input. - Examine indices for which the square root is an integer. - Boundedness alone does not guarantee that a sequence has one limiting value.

1. For \(n\ge2\), apply the Mean Value Theorem to \(f(x)=\cos(\pi\sqrt x)\) on \([n-1,n]\). Since \(|f'(x)|\le\frac{\pi}{2\sqrt{x}}\), \(|a_n|\le\frac{\pi}{2\sqrt{n-1}}\to0\). 2. At square indices, \(S_{m^2}=\cos(\pi m)=(-1)^m\). 3. Thus the subsequence \(S_{(2j)^2}\) equals \(1\), while \(S_{(2j+1)^2}\) equals \(-1\). 4. The partial sums do not have a limit, so the series diverges. Bounded partial sums and terms approaching zero are not sufficient for convergence.

The student is incorrect. Although \(a_n\to0\) and \((S_n)\) is bounded, the series diverges because its partial sums have subsequences equal to \(1\) and \(-1\).

- Look for a constant value that remains unchanged by the recurrence. - Measure each partial sum's distance from that constant. - Recover an individual term from two consecutive partial sums.

1. Subtract \(3\) from the recurrence to obtain \(S_n-3=\frac{n-1}{n}(S_{n-1}-3)\). 2. Repeated substitution gives \(S_n-3=\frac{1}{n}(S_1-3)=-\frac{3}{n}\), so \(S_n=3-\frac{3}{n}\). 3. Therefore \(S_n\to3\), and the series converges to \(3\). 4. For \(n\ge2\), \(a_n=S_n-S_{n-1}=\frac{3}{n(n-1)}\).

a) \(S_n=3-\frac{3}{n}\). b) The series converges to \(3\). c) \(a_n=\frac{3}{n(n-1)}\) for \(n\ge2\).

- Compare a partial sum of the grouped series with a partial sum of the original series. - For the converse, handle even and odd partial sums separately. - Identify the extra term present in an odd partial sum and use the additional condition on it.

1. Let \(S_n=\sum_{k=1}^{n}a_k\) and \(T_m=\sum_{k=1}^{m}b_k\). Grouping the first \(2m\) terms gives \(T_m=S_{2m}\). 2. For part a, \(S_n\to L\) implies its even subsequence \(S_{2m}\to L\), so \(T_m\to L\). 3. For part b, \(T_m\to L\) gives \(S_{2m}\to L\). Also \(S_{2m+1}=S_{2m}+a_{2m+1}\to L+0=L\). 4. The even and odd partial sums both approach \(L\), so \(S_n\to L\).

a) The grouped series has partial sums \(T_m=S_{2m}\), so it converges to \(L\). b) With \(a_n\to0\), both \(S_{2m}\) and \(S_{2m+1}\) approach \(L\), so the original series converges to \(L\).

- Rewrite the difference between the average and the proposed limit as an average of differences. - Separate the fixed initial portion from the long tail. - The influence of finitely many early terms becomes small when divided by a large index.

1. Fix \(\varepsilon>0\). Choose \(N\) so that \(|c_k-L|<\frac{\varepsilon}{2}\) for every \(k\ge N\). 2. Let \(C=\sum_{k=1}^{N-1}|c_k-L|\). For \(n\ge N\), \(|S_n-L|\le\frac{C}{n}+\frac{1}{n}\sum_{k=N}^{n}|c_k-L|\). 3. The second term is less than \(\frac{\varepsilon}{2}\). Choose \(n\) large enough that \(\frac{C}{n}<\frac{\varepsilon}{2}\). 4. Then \(|S_n-L|<\varepsilon\), so \(S_n\to L\) and the series converges to \(L\).

The series converges to \(L\).

- A convergent subsequence need not control values between its indices. - Build a partial-sum sequence that behaves differently away from the selected indices. - Any prescribed partial-sum sequence determines terms through consecutive differences.

1. Define \(S_n=0\) when \(n\) is a power of \(2\), and \(S_n=1\) otherwise. 2. Then \(S_{2^k}=0\) for every \(k\), so the stated subsequence converges to \(0\). 3. The full sequence does not converge because it also has infinitely many values equal to \(1\), for example \(S_{2^k+1}=1\). 4. Define \(a_1=S_1\) and \(a_n=S_n-S_{n-1}\) for \(n\ge2\). These terms produce exactly the prescribed partial sums, so the resulting series diverges.

The claim is false. One counterexample has \(S_n=0\) at powers of \(2\) and \(S_n=1\) otherwise, with \(a_1=S_1\) and \(a_n=S_n-S_{n-1}\).

- What fixed positive value appears to be approached by the recurrence? - Can you establish both a one-sided bound and a consistent direction of movement? - After convergence is known, what equation must the limiting value satisfy?

1. Since \(S_1=3>\sqrt6\), suppose \(S_n\ge\sqrt6\). Then \(S_{n+1}-\sqrt6=\frac{(S_n-\sqrt6)^2}{2S_n}\ge0\), so \(S_{n+1}\ge\sqrt6\). 2. Also \(S_{n+1}-S_n=\frac{6-S_n^2}{2S_n}\le0\), so the partial sums are nonincreasing. 3. Thus \((S_n)\) is bounded below and nonincreasing, so it converges to some \(L>0\). 4. Taking limits in the recurrence gives \(L=\frac12(L+6/L)\), so \(L^2=6\). 5. Because \(L>0\), \(L=\sqrt6\). Therefore the series converges to \(\sqrt6\).

The series converges to \(\sqrt6\).

- Track what happens to the block label as the index grows. - A term is zero whenever two consecutive partial sums are equal. - Nonzero changes occur only when the index enters a new block.

1. As \(n\to\infty\), the block index \(k\to\infty\), so \(S_n=1+\frac{1}{k+1}\to1\). The series converges to \(1\). 2. Within each block \(2^k\le n<2^{k+1}\), the partial sum is constant, so \(a_n=0\) except when a new block begins. 3. At \(n=2^k\) with \(k\ge1\), \(a_{2^k}=\left(1+\frac{1}{k+1}\right)-\left(1+\frac{1}{k}\right)=-\frac{1}{k(k+1)}\).

a) The series converges to \(1\). b) For \(n\ge2\), the only nonzero terms occur at \(n=2^k\) with \(k\ge1\), and \(a_{2^k}=-\frac{1}{k(k+1)}\).

- Locate an arbitrary large index between two consecutive square indices. - Compare its partial sum with the partial sum at the start of that square-index block. - Use the triangle inequality to combine the two available types of control.

1. For each sufficiently large index \(k\), choose \(m\) such that \(m^2\le k<(m+1)^2\). 2. The triangle inequality gives \(|S_k-L|\le|S_k-S_{m^2}|+|S_{m^2}-L|\). 3. The first term is at most \(\frac{1}{m}\), and the second approaches \(0\) because \(S_{m^2}\to L\). 4. As \(k\to\infty\), the associated \(m\to\infty\), so both terms approach \(0\). Hence \(S_k\to L\).

The series converges to \(L\).