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I tried to find a closed form for the series $$ \sum_{n=2}^\infty \frac{1}{n \sqrt{n^2-1}} $$ I got another form for the series by using the known series $$\frac{1}{\sqrt{n^2-1}}=\frac{1}{n\sqrt{1-n^{-2}}}=\sum_{k=0}^\infty \binom{2k}{k} \frac{1}{4^kn^{2k+1}} , \forall n>1$$ So $$ \sum_{n=2}^\infty \frac{1}{n \sqrt{n^2-1}}=\sum_{k=0}^\infty\binom{2k}{k} 2^{-2k} \left(\zeta(2k+2)-1 \right) $$ if there is no closed form can we get its integral representation ?

EDITION

here we have $$ \binom{2k}{k} 2^{-2k}=\frac{2}{\pi}\int_0^\infty \frac{1}{(x^2+1)^{k+1}} dx$$ then $$ \sum_{n=2}^\infty \frac{1}{n \sqrt{n^2-1}}=\frac{2}{\pi} \int_0^\infty \sum_{k=1}^\infty \frac{\zeta(2k)-1}{(x^2+1)^{k}} dx $$ and here we have the generation function of $\zeta(2n)$ $$ \sum_{n=1}^\infty \zeta(2n) x^n = \frac{1}{2} -\frac{\pi \sqrt{x}}{2} \cot(\pi \sqrt{x}) $$ finally $$ \sum_{n=2}^\infty \frac{1}{n \sqrt{n^2-1}}=\frac{1}{\pi} \int_0^\infty \left(1-\frac{2}{x^2}-\frac{\pi}{\sqrt{x^2+1}} \cot \left(\frac{\pi}{\sqrt{x^2+1}} \right) \right) dx$$

A rural reader
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Faoler
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2 Answers2

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Not an answer:

$$\sum _{n=2}^{\infty } \frac{1}{n \sqrt{n^2-1}}=\\\int_0^{\infty } \left(\mathcal{L}_n^{-1}\left[\frac{1}{n \sqrt{n^2-1}}\right](x)\right) \sum _{s=2}^{\infty } \exp (-s x) \, dx=\\\int_0^{\infty } \frac{\pi x (I_0(x) \pmb{L}_{-1}(x)-I_1(x) \pmb{L}_0(x)) e^{-x}}{2 \left(-1+e^x\right)} \, dx\approx0.69422401996922706=\sum _{m=0}^{\infty } \frac{(-1)^m \sqrt{\pi } (-1+\zeta (2+2 m))}{\Gamma \left(\frac{1}{2}-m\right) \Gamma (1+m)}$$

Where: $I_0(x)$ is modified Bessel function of the first kind, $\pmb{L}_0(x)$ is modified Struve function.

$\mathcal{L}_n^{-1}$ is Inverse Laplace Transform.

UPDATE:

Using:$$\int_0^1 \frac{1}{\left(n^2-x\right) \sqrt{x (1-x)} \pi } \, dx=\frac{1}{n \sqrt{n^2-1}}$$ we have: $$\int_0^1 \left(\sum _{n=2}^{\infty } \frac{1}{\left(n^2-x\right) \sqrt{x (1-x)} \pi }\right) \, dx=\int_0^1 -\frac{-1+3 x-\pi (-1+x) \sqrt{x} \cot \left(\pi \sqrt{x}\right)}{2 \pi (-((-1+x) x))^{3/2}} \, dx$$

Using:

$$\int_0^{\pi } \frac{2 x}{\pi ^2 \left(n^2-\cos ^2(x)\right)} \, dx=\frac{1}{n \sqrt{n^2-1}}$$

we have:

$$\int_0^{\pi } \left(\sum _{n=2}^{\infty } \frac{2 x}{\pi ^2 \left(n^2-\cos ^2(x)\right)}\right) \, dx=\\\int_0^{\pi } \left(-\frac{3 x \csc ^2(x)}{2 \pi ^2}-\frac{x \cot (\pi \cos (x)) \sec (x)}{\pi }+\frac{3 x \sec ^2(x)}{2 \pi ^2}-\frac{x \csc ^2(x) \sec ^2(x)}{2 \pi ^2}\right) \, dx$$

Mariusz Iwaniuk
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Here are a few more integral representations. Firstly, one can use the complex integral representation for $\binom nk$ and the generating function of $\zeta(2n)$ from the question:

$$\sum_{n=2}^\infty \frac1{n\sqrt{n^2-1}}= \sum_{n=0}^\infty\binom{2n}n2^{-2n} (\zeta(2n+2)-1)=\frac1\pi\int_{-\pi}^\pi\sec^2\left(\frac t2\right)\sum_{n=1}^\infty\cos^{2n}\left(\frac t2\right)(\zeta(2n)-1)dt$$

to find:

$$\boxed{\sum_{n=2}^\infty \frac1{n\sqrt{n^2-1}}=\int_0^\frac\pi2\frac1\pi\left(\sec^2(t)-2\csc^2(t)\right)-\cot(\pi\cos(t))\sec(t)dt}$$

shown here. An integral representation was given in the comment. To find it, use the inverse Laplace transform to get Bessel I:

$$\int_0^\infty e^{-st}I_0(t)dt=\frac1{\sqrt{s^2-1}}\implies\boxed{\sum_{n=2}^\infty \frac1{n\sqrt{n^2-1}}=-\int_0^\infty I_0(t)\left(e^{-t}+\ln\left(1-e^{-t}\right)\right)dt}$$

shown here. Finally, using $I_0(z)=\frac1\pi\int_0^\pi e^{z\cos(t)}dt$ gives the harmonic numbers:

$$\boxed{\sum_{n=2}^\infty \frac1{n\sqrt{n^2-1}}=\int_{-1}^1\frac{1-\pi\cot(\pi t)-H_{t-2}}{\pi t\sqrt{1-t^2}}dt}$$

as shown here. These so far do not seem to have a closed form.

Тyma Gaidash
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