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How does one calculate $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$$

if given this information: $$\sum_{n=1}^{\infty} \frac1{n^2} = \frac{\pi^2}{6}.$$

How does one account for the $(-1)^{n+1}$ in the first series?

Note: in this earlier post, user David Hodden (thanks) pointed out that the following simplification can be made:

$$\sum_{n=1}^{\infty} (-1)^{n+1}\frac1{n^2} = \sum_{n=1}^{\infty} \frac1{n^2} -2 \cdot \frac14 \sum_{n=1}^{\infty} \frac1{n^2} \\ = \frac{\pi^2}{12}$$

What I do not understand is where this simplification came from. Why can the alternating series be broken up into two non-alternating series?

Community
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Leo K.
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  • The information given tells you (in addition to the value) that the alternating series is absolutely convergent. This gives you greater freedom to manipulate it. – Henry Dec 04 '14 at 21:43

2 Answers2

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Notice that

$$\frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \ ... \ = \ \frac{1}{1^2} + \left( \frac{1}{2^2} - 2 \frac{1}{2^2}\right) + \frac{1}{3^2} + \left( \frac{1}{4^2} - 2 \frac{1}{4^2}\right) + \ ...$$

See where this is going?

Simon S
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There are two kind of terms in series $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$: terms with sign $+$ and terms with sign $-$. Note that:

$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\cdots$$

So $+$ occurs with terms for odd $n$ and $-$: for even terms. So let's divide this sum into two sums: for even and odd $n$:

$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}=\sum_{n=1 \text{n is odd}}^{\infty} \frac{(-1)^{n+1}}{n^2}+\sum_{n=1 \text{n is even}}^{\infty} \frac{(-1)^{n+1}}{n^2}$$

You know that $-$ occurs for odd $n$ and $+$ for even, so you can write down this equality in different way:

$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}=-\sum_{n=1 \text{n is odd}}^{\infty} \frac{1}{n^2}+\sum_{n=1 \text{n is even}}^{\infty} \frac{1}{n^2}$$

Next you can write down $\sum_{n=1 \text{n is even}}^{\infty} \frac{1}{n^2}$ in different way:

$$\sum_{n=1 \text{n is even}}^{\infty} \frac{1}{n^2}=\sum_{n=1}^{\infty} \frac{1}{(2n)^2}=\frac{1}{4}\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{24}$$

On the other hand:

$$\sum_{n=1 \text{n is odd}}^{\infty} \frac{1}{n^2}=\sum_{n=1}^{\infty} \frac{1}{n^2}-\sum_{n=1 \text{n is even}}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}-\frac{1}{4}\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}-\frac{\pi^2}{24}$$

agha
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