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This question was in my exam:

Calculate the series: $$\sum^\infty_{n=1}\frac{(-1)^n}{n^2}$$.

I answered wrong and the teacher noted: "You should use dirichlet's theorem".

I know my question is a bit general,

but can you please explain me how should I have solved this sum?

Thanks in advance.

Billie
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    By solve do you finding the actual limit of this series, or just showing whether or not it is convergent? – Hayden Feb 26 '14 at 14:09
  • Dirichlet's theorem, as I understand it, is a test for convergence (that is easily applicable here). Computing the value of the stated series is difficult to do from scratch. – Umberto P. Feb 26 '14 at 14:12
  • Or is it Dirichlet's theorem from the theory of Fourier series? – Etienne Feb 26 '14 at 14:16
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    It does not look clear to me what "Dirichlet's theorem" your teacher is referring to (Dirichlet's test? Dirichlet's theorem on primes in arithmetic progressions?), neither I would say that this is a "linear algebra" question. – Jack D'Aurizio Feb 26 '14 at 14:21

2 Answers2

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By absolute convergence you can simply write: $$\sum_{n=1}^{n}\frac{(-1)^n}{n^2}=\sum_{n \text{ even}}\frac{1}{n^2}-\sum_{n\text{ odd}}\frac{1}{n^2}=2\cdot\sum_{n \text{ even}}\frac{1}{n^2}-\sum_{n\geq 1}\frac{1}{n^2}=\frac{2}{4}\sum_{n\geq 1}\frac{1}{n^2}-\sum_{n\geq 1}\frac{1}{n^2}$$ $$=-\frac{1}{2}\sum_{n\geq 1}\frac{1}{n^2}=-\frac{\zeta(2)}{2}=-\frac{\pi^2}{12}.$$ Have a look at this hot question, too.

Community
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Jack D'Aurizio
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Jack's answer is perfectly all right.

If you want to use Dirichlet's theorem from the theory of Fourier series, consider the $2\pi$-periodic function $f$ such that $f(x)=x^2$ on $[-\pi,\pi)$.

Etienne
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