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I selected this sum $\sum_{n=1}^{+\infty} \zeta(n)e^{-n^2}$ for evaluation, my weaker assumptions showed me that is convergent for this reason the term $e^{-n^2}$ vanish when $n \to +\infty $ yield to the result $ +\infty.0$ which almost is $0$ because $\zeta(n)$ for large $n$ is diverge, Wolfram alpha say that is a convergent series then what is it's partial sum and it's closed form?

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  • $\begingroup$ the zeta Riemann is bounded in the positive reals greater than $1$ because $\lim_{n\to\infty}\zeta(n)=1$, it doesnt diverges. $\endgroup$ – Masacroso Feb 23 '18 at 12:15
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    $\begingroup$ I doubt there's a closed form for the series and even less likely for partial sums. $\endgroup$ – Yuriy S Feb 23 '18 at 12:29
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    $\begingroup$ Should $\zeta(1)$ be here though? Maybe you meant to write the general term as $\zeta(n+1)$? Because $\zeta(1)$ diverges on its own $\endgroup$ – Yuriy S Feb 23 '18 at 12:36
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    $\begingroup$ $\sum_{n\geq 2}\zeta(n)e^{-n^2}$ is blatantly convergent (not just because $e^{-n^2}\to 0$, but because $e^{-n^2}\to 0$ really fast) and $\sum_{n\geq 1}\zeta(n) e^{-n^2}$ is blatantly ill-defined since the $\zeta$ function has a simple pole at $s=1$. $\endgroup$ – Jack D'Aurizio Feb 23 '18 at 18:02
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The given series diverges because of the first term, as $\zeta(1)$ represents harmonic series and harmonic series diverges.

I will consider a convergent series instead:

$$S=\sum_{n=1}^\infty \zeta(n+1) e^{-n^2}$$

No closed form is expected to exist, but there's an interesting way to rewrite the series.

First, we rewrite it as a double sum:

$$S=\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{k^{n+1}} e^{-n^2}$$

Now we consider the term with $k=1$ separately, because otherwise it would cause us problems lately. In any case it has a closed form in terms of Jacobi theta functions:

$$\sum_{n=1}^\infty e^{-n^2}=\frac{1}{2} \left(\vartheta _3\left(0,\frac{1}{e}\right)-1\right)$$

Now let's move on to the rest of the series:

$$S_2=\sum_{k=2}^\infty \sum_{n=1}^\infty \frac{1}{k^{n+1}} e^{-n^2}$$

There's no closed form for the series in the form $\sum_{n=1}^\infty e^{-n^2} x^n$ (as far as I know or Mathematica knows), but we can rewrite it.

The following might be missing some rigour, but it works. Let's represent the exponential part as an integral:

$$e^{-n^2}= \frac{1}{2 \sqrt{ \pi}} \int_{- \infty} ^\infty e^{-x^2/4+inx} dx$$

The imaginary part of the integral is zero, because it is the integral of an odd function over a symmetric inverval. But it is beneficial to write it in exponential form anyway.

Getting back to the series:

$$S_2=\frac{1}{2 \sqrt{ \pi}} \sum_{k=2}^\infty \frac{1}{k} \int_{- \infty} ^\infty e^{-x^2/4} \sum_{n=1}^\infty \frac{1}{k^n} e^{inx} dx$$

Because $|e^{inx}| \leq 1$ and $k \geq 2$ we can write the closed form for the geometric series under the integral:

$$S_2=\frac{1}{2 \sqrt{ \pi}} \sum_{k=2}^\infty \frac{1}{k} \int_{- \infty} ^\infty e^{-x^2/4}\frac{ e^{ix}}{k-e^{ix}} dx$$

We can get rid of the imaginary part, which is again an odd function (as it should be):

$$\frac{ e^{ix}}{k-e^{ix}}=\frac{k \cos x-1+i k \sin x}{k^2-2k \cos x+1}$$

Now we can rewrite the series as a series of real valued integrals:

$$S_2=\frac{1}{2 \sqrt{ \pi}} \sum_{k=2}^\infty \frac{1}{k} \int_{- \infty} ^\infty e^{-x^2/4} \frac{k \cos x-1}{k^2-2k \cos x+1} dx$$

The series under the integral has a closed form too, in terms of digamma functions:

$$\sum_{k=2}^\infty \frac{1}{k} \frac{k \cos x-1}{k^2-2k \cos x+1}=1- \gamma-\frac{1}{2} \left( \psi (2- \cos x+ i \sin x) +\psi (2- \cos x- i \sin x) \right)$$

Where $\gamma$ is Euler-Mascheroni constant and the value is real, because the arguments are conjugate.

So we can write:

$$S_2=1- \gamma-\frac{1}{4 \sqrt{ \pi}} \int_{- \infty} ^\infty e^{-x^2/4} \left( \psi (2- \cos x+ i \sin x) +\psi (2- \cos x- i \sin x) \right) dx$$

Naming the integral (which I believe has no closed form) and getting its numerical value from Mathematica, we have:

$$I=\int_{- \infty} ^\infty e^{-x^2/4} \left( \psi (2- \cos x+ i \sin x) +\psi (2- \cos x- i \sin x) \right) dx= \\ =1.2890375247789216487\dots$$

Getting back to the full series we obtain:

$$S=\frac{1}{2} \left(\vartheta _3\left(0,\frac{1}{e}\right)+1\right)-\gamma-\frac{I}{4 \sqrt{ \pi}}=0.62728755144118062\dots$$

This is the same as numerical value Mathematica obtains for the original series.


We can also use the integral form of the digamma function to rewrite $I$ as a double integral and get a curious formula:

$$S=\frac{1}{2} \left(\vartheta _3\left(0,\frac{1}{e}\right)+1\right)-\frac{1}{2 \sqrt{ \pi}} \int_{- \infty} ^\infty \int_0^1 \frac{1-t^{1-\cos (x)} \cos (\log (t) \sin (x))}{1-t}~e^{-x^2/4}~ dt~dx$$

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  • $\begingroup$ Thanks for your answer but why wolfram alpha assumed that is convergent for n=1 to +infty ? $\endgroup$ – zeraoulia rafik Feb 23 '18 at 21:20
  • $\begingroup$ @zeraouliarafik, good question. Wolfram Alpha (and even Mathematica) sometimes doesn't work. You should use common sense to check your results. $\zeta(1)$ diverges, it is well known. $\endgroup$ – Yuriy S Feb 23 '18 at 21:23

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