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Show that \(\displaystyle\sum_{n\geq1}\frac{(-1)^{n-1}}{n(n+1)\cdots(n+k)}=\frac{2^k}{k!}\left(\ln2-\sum_{i=1}^{k}\frac{(1/2)^i}{i}\right)\) where \(k\) is a non negative integer and when \(k=0\) the second sum is to be interpreted as the sum over the empty set, i.e. \(=0\).

Show that

\(\displaystyle \int_{-\infty }^{\infty }\int_{-\infty }^{\infty }{\rm sign}(x)\,{\rm sign}(y)e^{-\frac{x^2+y^2}{2}}\sin (xy)\,dx\,dy=2\sqrt{2}\log (1+\sqrt{2})\).

Hello. Since \(\zeta(1/2)\) cannot be expressed using other known mathematical constants, we pretty much have a matter of definition here. The values of \(\zeta(s)\) for \(\Re(s)\leq1\) , \(s\neq1\) are defined via analytic continuation. This can be done for example using the Euler Mac Laurin Summat...

Let's take it a little bit further:

If \(A_n=\sqrt[n^2]{\binom{n+1}{1}\binom{n+2}{2}\cdots\binom{n+n}{n}}\), show that

\[\displaystyle{A_n=\frac{4}{\sqrt{e}}\left(1-\frac{\ln n}{2n}+\frac{\ln(4/\sqrt{2\pi})}{n}+\frac{\ln^2n}{8n^2}+\mathcal O\left(\frac{\ln n}{n^2}\right)\right)}\]

Thank you. We can also write

\(\displaystyle s_n=\sum_{k=0}^{n}\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+2}=H_{4n+3}-\frac{1}{2}\left(H_{2n+1}+H_{n+1}\right)\to\frac{3\ln2}{2}\).

Two more approaches can be seen here .

Calculate the following:

\(\displaystyle1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\cdots\).

Good morning. The problem is from here The solution I had sent: (It uses DCT too) \begin{align}\sum_{n\geq0}(-1)^n\sum_{k\geq1}(-1)^{k-1}\frac{x^{n+k}}{n+k}&=\sum_{n\geq0}\sum_{k\geq1}(-1)^{n+k-1}\frac{x^{n+k}}{n+k}\notag \\ &\stackrel{m:=n+k}{=}\sum_{n\geq0}\sum_{m\geq n+1}(-1)^{m-1}\frac{x...

For \(x\in(-1,1]\), calculate \(\displaystyle\sum_{n\geq0}(-1)^n\left(\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}+\frac{x^{n+3}}{n+3}-\cdots\right)\).

Proposed by Ovidiu Furdui.

Evaluate \(\displaystyle\sum_{n\geq1}\frac{4^n}{\binom{2n}{n}(4n^2-1)}\).

For \(k>0\) and \(n\geq0\) real numbers, calculate \(\displaystyle \int_{0}^{1}x^n\ln\left(\sqrt{1+x^k}-\sqrt{1-x^k}\right)\,dx\).

Proposed by Ovidiu Furdui.