The skill being tested here is the alternating series error bound, which provides a way to estimate the error when approximating an infinite alternating series with a partial sum. For an alternating series that satisfies the conditions of the alternating series test—terms alternating in sign, decreasing in absolute value, and approaching zero—the error in using the partial sum S_n is less than the absolute value of the next term, $a_{n+1}$. This bound works because the remainder after n terms is bracketed between zero and the first omitted term, ensuring the actual sum lies within S_n ± $|a_{n+1}$|. In this case, for the series ∑ $(-1)^{k+1}$$/k^3$ approximated by S_10, the next term is $1/11^3$, so the error is at most $1/11^3$. A tempting distractor like ≤ $1/10^3$ fails because it uses the last included term instead of the next one, which underestimates the bound since the error is actually smaller than the next term, not the previous. A transferable strategy for error bounds in alternating series is to always identify the first omitted term after verifying the series meets the convergence criteria.

This question tests the alternating series error bound for $\sum_{n=1}^{\infty}(-1)^{n}\frac{3}{\sqrt{n}}$ using 50 terms. When we stop at $n=50$, the first omitted term is at $n=51$. The absolute value of this term is $\left|(-1)^{51}\frac{3}{\sqrt{51}}\right| = \frac{3}{\sqrt{51}}$. Choice E, $\left|\frac{3}{\sqrt{51}}-\frac{3}{\sqrt{50}}\right|$, incorrectly suggests using the difference between consecutive terms rather than the next term itself. The key principle for alternating series error bounds is that the error is bounded by the absolute value of the first term not included in your partial sum.

The skill being tested here is the alternating series error bound, which provides an upper limit on the approximation error for convergent alternating series. For $∑_{k=0}$^∞ $(-1)^k$ $(4/(k+2)^2$), terms alternate with decreasing sizes to zero. After n terms, meaning up to k=n, the error bound is $4/(n+3)^2$, the magnitude of the term at k=n+1. This bound is valid as the remainder is an alternating series itself, with absolute value less than its first term due to monotonic decrease. A tempting distractor like $4/(n+2)^2$ fails by using the last included term instead of the omitted one. Always identify the general term and ensure the error bound is the absolute value of the (n+1)th term for transferable strategy across alternating series.

This problem asks for the alternating series error bound when approximating $\sum_{n=1}^{\infty}(-1)^{n}\frac{5}{\sqrt{n}}$ by $S_{25}$. The Alternating Series Error Bound theorem tells us that when we truncate a convergent alternating series at the nth term, the error is at most the absolute value of the (n+1)th term. Since we're using $S_{25}$ (the sum of the first 25 terms), the error is bounded by the 26th term: $|a_{26}| = \frac{5}{\sqrt{26}}$. Choice B ($\le \frac{5}{\sqrt{25}}$) incorrectly uses the last included term rather than the first omitted term, a common mistake. Remember that the error bound for alternating series is always the absolute value of the first term you leave out, not the last term you include.

This problem involves finding the alternating series error bound for a series with a more complex general term. For $S=\sum_{k=2}^{\infty}(-1)^k\frac{3}{k\ln k}$, when approximating by $S_5$, we include terms for $k=2,3,4,5$. The error bound equals the absolute value of the first omitted term, which is at $k=6$: $|(-1)^6\frac{3}{6\ln 6}| = \frac{3}{6\ln 6}$. Choice B incorrectly uses $k=5$ (the last included term) rather than $k=6$ (the first excluded term). When a series starts at an index other than 0 or 1, carefully track which term is the first one you're not including in your partial sum.

The skill being tested here is the alternating series error bound, which provides a way to estimate the error when approximating an infinite alternating series with a partial sum. For an alternating series that satisfies the conditions of the alternating series test—terms alternating in sign, decreasing in absolute value, and approaching zero—the error in using the partial sum S_n is less than the absolute value of the next term, $a_{n+1}$. This bound works because the remainder after n terms is bracketed between zero and the first omitted term, ensuring the actual sum lies within S_n ± $|a_{n+1}$|. In this case, for the series ∑ $(-1)^k$ * 2/(k+1) approximated by S_6, the next term is 2/8, so the error is at most 2/8. A tempting distractor like ≤ 2/7 fails because it uses the last included term instead of the next one, which underestimates the bound since the error is actually smaller than the next term, not the previous. A transferable strategy for error bounds in alternating series is to always identify the first omitted term after verifying the series meets the convergence criteria.

This problem tests the alternating series error bound, which states that the error when truncating an alternating series is at most the absolute value of the first omitted term. Since we're using the first 12 terms of $\sum_{k=1}^{\infty}(-1)^{k+1}\frac{1}{k^3}$, the first omitted term is the 13th term, which has $k=13$. The absolute value of this term is $\left|(-1)^{13+1}\frac{1}{13^3}\right| = \frac{1}{13^3}$. Choice E, $\frac{1}{13^3-12^3}$, incorrectly attempts to use the difference between consecutive denominators rather than the next term itself. For alternating series satisfying the conditions (decreasing terms approaching zero), always use the absolute value of the first omitted term as your error bound.

The skill being tested here is the alternating series error bound, which provides a way to estimate the error when approximating an infinite alternating series with a partial sum. For an alternating series that satisfies the conditions of the alternating series test—terms alternating in sign, decreasing in absolute value, and approaching zero—the error in using the partial sum S_n is less than the absolute value of the next term, $a_{n+1}$. This bound works because the remainder after n terms is bracketed between zero and the first omitted term, ensuring the actual sum lies within S_n ± $|a_{n+1}$|. In this case, for the series ∑ $(-1)^{k-1}$ * $3/(k^2$+1) approximated by S_8, the next term is $3/(9^2$+1), so the error is at most $3/(9^2$+1). A tempting distractor like ≤ $3/(8^2$+1) fails because it uses the last included term instead of the next one, which underestimates the bound since the error is actually smaller than the next term, not the previous. A transferable strategy for error bounds in alternating series is to always identify the first omitted term after verifying the series meets the convergence criteria.

This question applies the alternating series error bound to $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{6}{\ln(n+1)}$ after 9 terms. Using terms from $n=1$ to $n=9$, the first omitted term is at $n=10$. The absolute value of this term is $\left|(-1)^{10-1}\frac{6}{\ln(10+1)}\right| = \frac{6}{\ln(11)}$. Choice E, $\left|\frac{6}{\ln(11)}-\frac{6}{\ln(10)}\right|$, represents a common error of using the difference between consecutive terms instead of the next term itself. For alternating series satisfying the required conditions, the error is always bounded by the absolute value of the first omitted term.

This question tests the alternating series error bound for $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{8}{3n-1}$ after 25 terms. Using terms from $n=1$ to $n=25$, the first omitted term occurs at $n=26$. The absolute value of this term is $\left|(-1)^{26-1}\frac{8}{3(26)-1}\right| = \frac{8}{77}$, which equals $\frac{8}{3\cdot 26-1}$. Choice E, $\frac{8}{(3\cdot 26-1)-(3\cdot 25-1)}$, incorrectly uses the difference between consecutive denominators, which would give $\frac{8}{3} = \frac{8}{77-74}$, a much larger value. Always remember that the alternating series error bound is simply the absolute value of the next term in the series.