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.

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 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 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.

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.

This problem involves finding the alternating series error bound for a series with square root terms. For convergent alternating series where terms decrease in absolute value, the error from using n terms is bounded by the (n+1)st term's absolute value. Since we use $S_{49}$ (sum of first 49 terms), the error is bounded by the 50th term. The 50th term is $(-1)^{50-1} \cdot \frac{1}{\sqrt{50}} = -\frac{1}{\sqrt{50}}$, so the error bound is $\frac{1}{\sqrt{50}}$. Students might choose $\frac{1}{\sqrt{49}}$ thinking it's the last included term, but error bounds use the first excluded term. Remember: alternating series error bound = $|\text{first omitted term}|$.

This question involves finding alternating series error bounds for geometric-like series. When approximating a convergent alternating series with decreasing terms using n terms, the error is bounded by the $(n+1)$st term's absolute value. Using $S_8$ means summing the first 8 terms, so the error bound is the 9th term. The 9th term is $(-1)^{9+1} \cdot \frac{4}{3^9} = \frac{4}{3^9}$. A common error is using $\frac{4}{3^8}$ (the 8th term) instead of the 9th term. Remember that alternating series error bounds always use the first omitted term, not the last included term.

This problem asks for the alternating series error bound when approximating $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{6}{3n+2}$ by $S_{18}$. The Alternating Series Error Bound states that the error in truncating a convergent alternating series at the nth term is bounded by the absolute value of the (n+1)th term. Using $S_{18}$ means we include terms through $n=18$, so the error is bounded by the 19th term: $|a_{19}| = \frac{6}{3(19)+2} = \frac{6}{57+2} = \frac{6}{59} = \frac{6}{3 \cdot 19 + 2}$. Choice A ($\le \frac{6}{3 \cdot 18+2}$) represents the last included term rather than the first omitted one, a common conceptual error. Remember that the alternating series error bound always uses the first term you don't include in your approximation.