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Thank you for the answer Demetres. The problem is from here.

Show that $$\sum_{k=1}^{n-1}\csc\left(\frac{k\pi}{n}\right)=\frac{2}{\pi}n\ln n+\left(\frac{2\gamma}{\pi}-\frac{2\ln(\pi/2)}{\pi}\right)n+\mathcal O(1)\,.$$

We have \(\begin{align*}\displaystyle\sum_{n\geq2}\frac{(-1)^n\zeta(n)}{n(n+1)}=\sum_{n\geq2}\frac{(-1)^n\zeta(n)}{n}-\sum_{n\geq2}\frac{(-1)^n\zeta(n)}{n+1}:=A-B\end{align*}\). Now from the Taylor series of \(\ln(1-x)\) it is \(\begin{align}\label{1}\displaystyle\sum_{n\geq2}\frac{x^n}{n}&=-\ln...

Is there a "direct" way computing of $\displaystyle\sum_{n=2}^\infty \frac{(-1)^n\,\zeta(n)}{n\,(n+1)}$ ?

A nice relevant problem, proposed by Ovidiu Furdui, that is still open for submitting solutions is H722 here
http://www.fq.math.ca/Problems/2012AugustAdvanced.pdf

Set \(S_N:=\sum_{n=1}^{N}(-1)^n\left(\frac{1}{\sqrt{n}}+\frac{(-1)^{n-1}}{n}\right)\), so that \(S_N=-H_N+\sum_{n=1}^{N}\frac{(-1)^n}{\sqrt{n}}\). By Dirichlet's criterion \(\sum_{n\geq1}\frac{(-1)^n}{\sqrt{n}}\) converges, so, since \(H_N=\ln N+\mathcal O(1)\) we have \(S_N=-\ln N+\mathcal O(1)\) t...

Another solution: We set \(\displaystyle{a_n:=-\int_{0}^{+\infty}\frac{\ln x}{(1+x)^n}\,dx}\), multiply with \(y^n\) and sum for \(n\geq2\). For \(y\in(-1,0)\cup(0,1)\) we will have: \[\begin{aligned}\sum_{n\geq2}a_ny^n&=-\sum_{n\geq2}\left(\int_{0}^{+\infty}\frac{\ln x}{(1+x)^n}\,dx\right)y^n \...

Thank you for your solution Demetres. I don't have a combinatorial proof either. Here is the solution I had given for this one on http://www.mathematica.gr.: At first we have \(\begin{aligned}\displaystyle\sum_{k=0}^{n-1}(-1)^{n-k-1}\dfrac{(n+k)!}{(k!)^2(n-k-1)!}&=\sum_{k=0}^{n-1}(-1)^{n-k-1}(n-...

Let \(a_n\) be defined by \(\displaystyle{a_n=n(n-1)a_{n-1}+\frac{n(n-1)^2}{2}a_{n-2}}\) for \(n\geq3\) and \(a_{1}=0,\;a_2=1\). \(\displaystyle{1)}\) Show that \(\displaystyle{\lim_{n\to+\infty}\frac{e^{2n}a_n}{n^{2n+1/2}}=2\sqrt{\frac{\pi}{e}}}\) and \(\displaystyle{2)}\) compute \(\displaystyle{\...

Evaluate the following limits if they exist

1) \(\displaystyle \lim_{s\to+\infty}\frac{1}{\ln s}\int_{0}^{+\infty}\frac{e^{-\frac{x}{s}-\frac{1}{x}}}{x}\,dx\)



2) \(\displaystyle \lim_{s\to+\infty}\left(\int_{0}^{+\infty}\frac{e^{-\frac{x}{s}-\frac{1}{x}}}{x}\,dx-\ln s\right)\).