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?

  • $\begingroup$ 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
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 3
    $\begingroup$ $\sum_{n\geq 2}\zeta(n)e^{-n^2}$ 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

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:

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.