Find closed form of the following -
$$ \displaystyle \sum_{n=2}^{\infty}{\left(\frac{(n-1)\zeta(n)}{4n-1}\right)} $$
I don't know how to approach to it - Using the integral definition? I cannot use because I came to this result from there only.
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1As written, the series is trivially diverging. Maybe you want to replace $\zeta(n)$ with $\left(\zeta(n)-1\right)$. – Jack D'Aurizio Aug 27 '15 at 12:25
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Okay then replace $ \zeta(n) \rightarrow (\zeta(n) -1) $ and then find the answer. It might be that I read the problem wrongly. @JackD'Aurizio – Kartik Sharma Aug 28 '15 at 17:29
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One may observe that $$ \zeta(n)=1+\frac1{2^n}+\frac1{3^n}+\cdots,\quad n>1, $$ gives $$ \lim_{n \to \infty}\zeta(n)=1, $$ then
$$ \lim_{n \to \infty}{\left(\frac{(n-1)\zeta(n)}{4n-1}\right)}=\frac14 \times 1\neq0. $$
Your series, as written, is divergent.
Olivier Oloa
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Oh, I see. Should this also diverge - $$ \displaystyle \sum_{n=2}^{\infty}{\left(\frac{(n-1)\zeta(n)}{4n-1} - 1\right)} $$ ? – Kartik Sharma Aug 27 '15 at 09:51
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Actually I was working on the following summation $$\sum _{ s=1 }^{ \infty }{ \left( \frac { 1 }{ 4s-1 } \sum _{ n=0 }^{ \infty }{ \left( \frac { 1 }{ n+1 } \sum _{ k=0 }^{ n }{ \left( \left( \begin{matrix} n \ k \end{matrix} \right) \frac { { \left( -1 \right) }^{ k } }{ { \left( k+1 \right) }^{ s-1 } } \right) } \right) } -1 \right) }$$ and I am getting the sum as written above. It's my bad that I changed the problem. @Olivier Oloa – Kartik Sharma Aug 27 '15 at 09:55
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@KartikSharma Did you mean to write $\left(\dotsb-\dfrac14\right)$? He said that the limit was $\frac14$, so I'd expect you to subtract $\frac14$ to make the limit $0$. – Akiva Weinberger Aug 27 '15 at 09:55
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I'd expect to do as you said. But then that is not what I get from the original problem as stated above. Check out, please! – Kartik Sharma Aug 27 '15 at 09:57
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$$\sum_{n\geq2}\frac{\zeta\left(n\right)\left(n-1\right)}{4n-1}\geq\frac{1}{4}\sum_{n\geq2}\frac{\zeta\left(n\right)}{n}\geq\frac{1}{4}\sum_{n\geq2}\frac{1}{n}. $$
Marco Cantarini
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If you replace $\zeta(n)$ with $\zeta(n)-1$ (otherwise the series is trivially diverging, as shown by the other answers), since $\sum_{n\geq 2}\left(\zeta(n)-1\right)=1$ we just need to compute: $$ S = \sum_{n\geq 2}\frac{\zeta(n)-1}{4n-1}. \tag{1}$$ On the other hand, $$ \sum_{n\geq 2}\zeta(n)\, z^n = -z\left(\gamma+\psi(1-z)\right)\tag{2}$$ hence by replacing $z$ with $z^4$, then dividing by $z^2$: $$ \sum_{n\geq 2}\zeta(n)\,z^{4n-2} = -z^2(\gamma+\psi(1-z^4))\tag{3}$$ so: $$ \sum_{n\geq 2}\left(\zeta(n)-1\right)\,z^{4n-2} = -\gamma z^2-z^2\psi(1-z^4)-\frac{z^6}{1-z^4}\tag{4}$$ and: $$ S = -\frac{\gamma+1}{3}-\int_{0}^{1}x^2\,\psi(2-x^4)\,dx \tag{5}$$ but honestly I do not think the last integral has a nice closed form.
Jack D'Aurizio
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