How do I calculate the value of this series?

Question

I want to find the value to which this series converges $$\sum_{n=0}^\infty \frac{(-1)^n}{n^2+1}$$

I tried looking at the sequence of partial sums $$S_k = \sum_{n=0}^k \frac{(-1)^n}{n^2+1}$$ and I noticed that $$\frac{-1}{n^2+1} \leq \frac{(-1)^n}{n^2+1} \leq \frac{1}{n^2 +1}$$ and so I think that by the squeeze rule I can see ( I could have just noticed it by logic, but okay) that the terms converge to zero. How do I find the value of the original series though? I could only show that it converged

https://www.wolframalpha.com/input/?i=sum+((-1)%5En)%2F(n%5E2%2B1),+n%3D0+to+infinity, has a complicated converged value! — 2017-02-20
@stud_iisc that's trivial, but to derive a real representation is much more interesting. — 2017-02-20
@stud_iisc however I am asked to find the value of it! I tried on wolfram alpha as well, in my computer gave me the answer in terms of cosecants or such things.. it shouldn't be that hard, it's a second year university exercise — 2017-02-20
For example,$$\sum_{n=0}^\infty\frac{(-1)^n}{n^2+1}=\int_0^1\sum_{n=0}^\infty(-x)^{n^2}\ dx$$ — 2017-02-20
@stud_iisc Wolfram Alpha seems to have missed the $n=0$ term when I follow your link. I make the sum about $+0.636$ — 2017-02-20
If you were asked to give the value then perhaps you've already dealed with series from $\;-\infty\;$ to $\;\infty\;$ , Dirichlet series or stuff like that. Then (obsderve the index!) $$\sum_{n=-\infty}^\infty\frac{(-1)^n}{n^2+1}=\pi\,\text{csch}\,\pi $$ — 2017-02-20
@Henry Yes you are right. It's including $n=0$ term but is showing the summation from $n=1$ term. — 2017-02-20
@DonAntonio unluckily I haven't met the Dirichlet series. I just had a basic analysis module. This question however is in a partial differential equation exercise, precisely in the Fourier Series chapter. — 2017-02-20
@Euler_Salter I honestly can't tell you even how to try without more powerful tools, but I think Simply Beautiful Art has some ideas. At least there's something that could, perhaps, yield something interesting when applying to it Laplace Transform (as you're studying diff. eq's perhaps that's what is expected from you...) — 2017-02-20

user id:44121 294k · Accepted Answer

You may notice that: $$ \frac{1}{n^2+1} = \int_{0}^{+\infty}\sin(x)e^{-nx}\,dx\tag{1}$$ from which$^{(*)}$: $$ S=\sum_{n\geq 0}\frac{(-1)^n}{n^2+1}=1+\int_{0}^{+\infty}\sin(x)\sum_{n\geq 1}(-1)^n e^{-nx}\,dx \tag{2}$$ and: $$ S = 1-\int_{0}^{+\infty}\frac{\sin(x)}{e^x+1}\,dx =\color{red}{\frac{1}{2}\left(1+\frac{\pi}{\sinh\pi}\right)}\tag{3}$$ where the last equality follows from integration by parts and the residue theorem. The same can be proved by considering the Fourier cosine series of $\cosh(x)$ over the interval $(-\pi,\pi)$.

Yet another (Eulerian) approach. It is clearly enough to compute $\sum_{n\geq 1}\frac{1}{n^2+1}$ and $\sum_{n\geq 1}\frac{1}{4n^2+1}$.
From the Weierstrass product for the $\sinh$ function we have $$ \frac{\sinh(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1+\frac{z^2}{n^2}\right)\tag{4}$$ and by applying $\frac{d}{dz}\log(\cdot)$ to both sides: $$ -\frac{1}{z}+\pi\coth(\pi z) = \sum_{n\geq 1}\frac{2z}{z^2+n^2}\tag{5}$$ At last we just need to evaluate the LHS of $(5)$ at $z=1$ and $z=\frac{1}{2}$.

$(*)$ The exchange of $\sum$ and $\int$ is allowed by the absolute convergence of the series $\sum_{n\geq 0}\frac{(-1)^n}{n^2+1}$, the trivial inequality $\left|\sin(x)\right|\leq x$ and the dominated convergence theorem. For any $x>0$ we have $$ \sum_{n=1}^{N} e^{-nx}\leq \frac{1}{e^x-1} $$ and $\frac{x}{e^x-1}$ is a function belonging to $\mathcal{L}^1(\mathbb{R}^+)$, whose integral over $\mathbb{R}^+$ equals $\zeta(2)=\frac{\pi^2}{6}$.

Many ways to write the same thing. Was just in the middle of trying this, though it is the first time I've tackled a series like this. +1 — 2017-02-20
The Fourier cosine series comes out to my last comment on the main post, right? — 2017-02-20
@SimplyBeautifulArt: I am noticing it just now, but sure, it is a pretty standard technique in this case. — 2017-02-20
Another question I had for a longest time - how to evaluate series like this without complex analysis? — 2017-02-20
@YuriyS: $(4)$ can be proved through real-analytic tools only: it can be proved through Chebyshev polynomials. — 2017-02-20
Am I supposed to be able to solve this in a second year of mathematics? — 2017-02-20
@Euler_Salter: if you are aware of $(4)$ it is not a difficult problem. Otherwise, elements of Fourier or complex analysis do the job equally fine. If you know just the basics, it is not an easy question at all. — 2017-02-20
@JackD'Aurizio but how did you see the first equality? I mean, did you just went looking for a function or did you know it by heart? — 2017-02-20
@Euler_Salter: by experience. It is a useful identity for dealing with similar series, like $$\sum_{n,m\geq 1}\frac{1}{(n^2+m^2)^2}$$ — 2017-02-20
@JackD'Aurizio: I mean I know and can apply residue theorem and related complex analysis theorem, I know how to computer Fourier series (well just starting) and I know some basic analysis. But I could have never merged all of this together.. — 2017-02-20
(+1) The Eisenstein series for $\cot$ (and friends) deserves to be better-known... it seems to have fallen off the map in recent years! — 2017-02-21
@JackD'Aurizio: I fail to see how you know that you can exchange the sum with the integral. Maybe you should comment on that on your answer? — 2017-02-21
@Euler_Salter I first saw that integral when solving $$\int_0^\infty\frac{\sin(x)}x\ dx$$ — 2017-02-21
@MartinArgerami: $\sum_{n\geq 1}\frac{(-1)^n}{n^2+1}$ is absolutely convergent and the dominated convergence theorem applies since $\left|\sin x\right|\leq x$ and $\int_{0}^{+\infty}\frac{x}{e^x-1}\,dx = \zeta(2)$. — 2017-02-21
@JackD'Aurizio: fair enough, but I seriously doubt that the OP nor many readers know about dominated convergence; and then several of the few who do, will still struggle to find the dominating function. My guess is that most will simply think that the exchange is just fine. — 2017-02-21
@MartinArgerami I didn't know about the theorem indeed, I do know you always need to be careful from our course in Analysis though. — 2017-02-21
@JackD'Aurizio although might imply some work, would you be able to add some extra steps and considerations? — 2017-02-21

user id:221811 56k · Accepted Answer

Extend the sum $S$ to $$ 2S-1 = \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n^2+1}, $$ which we can get away with because the summand is even in $n$. We want to find something that gives these as residues, then use Cauchy's theorem, hoping the integral goes to zero. Therefore we take $$ \frac{\pi\csc{\pi z}}{z^2+1}: $$ this has poles at $z=n$ with residue $ (-1)^n/(n^2+1) $, and two others, at $\pm i$, with residues $ \frac{\pi \operatorname{csch}{(\pm \pi)}}{\mp 2i} $. Also, if we take the integral to be around a large square that between the poles, one can show using $ \lvert\sin{(x+iy)} \rvert^2 = \sin^2{x}+\sinh^2{y} $ and the corresponding one for cosine that the integrand is bounded on the border of the square by a multiple of $1/(x^2+y^2)$, and hence the integral tends to zero as we make the square larger and larger. Hence $$ \frac{1}{2\pi i}\left( 2S_k-1 - \frac{2\pi\operatorname{csch}{\pi}}{2} \right) = \int_{\square_k} \frac{\pi\csc{\pi z}}{z^2+1} \, dz \to 0 $$ as $k \to \infty$, and then you can rearrange to get the answer.

The general idea is that $\pi\cot{\pi z}$ has residue $1$ at every integer $n$, while $\csc{\pi z}$ has residue $(-1)^n$ at every integer $n$. The same technique works on any sum of the form $1/p(n^2)$ (or $(-1)^n/p(n^2)$) where $p$ is a nonconstant polynomial, although you do have to find the roots.

I was just in the middle of preparing this way forward as a solution. (+1) for the faster typing! — 2017-02-20

How do I calculate the value of this series?

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