Use [math: ∑n=1∞1n2=π26]\displaystyle \sum\limits_{n=1}^ \infty \frac{1}{n^2} = \frac{\pi{^2}}{6} to find the sum of the series [math: ∑n=1∞(−1)n+1n2]\displaystyle \sum\limits_{n=1}^ \infty \frac{(-1)^{n+1}}{n^2} .

Question

BlackTom AI · Accepted Answer

Let's begin by restating the problem in our own words: we know the classic series sum from n=1 to infinity of 1/n^2 equals pi^2/6, and we are asked to find the sum of the alternating series sum from n=1 to infinity of (-1)^(n+1)/n^2.

Option a claims the alternating sum equals -pi^2/6. This would be the value of the non-alternating sum if all terms carried the same sign and there were no alternating behavior, but the (-1)^(n+1) factor makes successive terms alternate in sign, so the sum cannot simply be the negative of the positive-terms series. Thus a is inconsistent with the alternating structure.

Option b suggests the sum equals 0. An alternating sum can sometimes cancel to zero, but in this case the terms decrease in magnitude as 1/n^2 and the sign alternates; however, the first term is positive (n=1 gives +1), so the entire infinite sum cannot be zero because the partial sums are positive at least for the initial terms and converge to a positive value. Therefore b is incorrect.

Option c asserts the sum equals pi^2/12. This aligns with the known relation for the alternating series of even powers: sum_{n=1}^∞ (-1)^{n+1}/n^2 = (1 - 2^{1-2}) * zeta(2) = (1 - 1/2) * (pi^2/6) = (1/2) * (pi^2/6) = pi^2/12. This uses the general Dirichlet eta function relation or the more familiar formula that the alternating zeta value is (1 - 2^{1-s}) times zeta(s) for s=2. Thus c matches the correct derivation.

Option d claims the sum equals pi^2/8. This value does not arise from the standard relation between the alternating and non-alternating zeta values at s=2, since (1 - 2^{1-2}) equals 1/2, not 3/4 or any factor producing pi^2/8. Hence d is incorrect.

In summary, while a and b misinterpret the effect of alternating signs on the sum, and d uses an incorrect normalization, option c is the one that coherently follows from the known connection between the ordinary zeta value zeta(2) = pi^2/6 and the alternating sum involving (-1)^{n+1}/n^2.

Use [math: ∑n=1∞1n2=π26]\displaystyle \sum\limits_{n=1}^ \infty \frac{1}{n^2} = \frac{\pi{^2}}{6} to find the sum of the series [math: ∑n=1∞(−1)n+1n2]\displaystyle \sum\limits_{n=1}^ \infty \frac{(-1)^{n+1}}{n^2} .单项选择题

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