 Carole Juarez

2022-02-22

How do we find
$S=\sum _{k=1}^{\mathrm{\infty }}\left[\frac{1}{2k}-\mathrm{log}\left(1+\frac{1}{2k}\right)\right]$ Donald Erickson

Observe that, by absolute convergence:
$\sum _{k=1}^{\mathrm{\infty }}\left(\frac{1}{2k}-\mathrm{log}\left(1+\frac{1}{2k}\right)\right)=\sum _{k=1}^{\mathrm{\infty }}\frac{1+{\left(-1\right)}^{k}}{2}\left(\frac{1}{k}-\mathrm{log}\left(1+\frac{1}{k}\right)\right)$
$=\frac{1}{2}\sum _{k=1}^{\mathrm{\infty }}\left(\frac{1}{k}-\mathrm{log}\left(1+\frac{1}{k}\right)\right)+\frac{1}{2}\sum _{k=1}^{\mathrm{\infty }}{\left(-1\right)}^{k-1}\left(\frac{1}{k}-\mathrm{log}\left(1+\frac{1}{k}\right)\right)$
$=\frac{\gamma }{2}+\frac{1}{2}\sum _{k=1}^{\mathrm{\infty }}{\left(-1\right)}^{k-1}\left(\frac{1}{k}-\mathrm{log}\left(1+\frac{1}{k}\right)\right)$
Then, as N is great, write
$\sum _{k=1}^{N}{\left(-1\right)}^{k-1}\left(\frac{1}{k}-\mathrm{log}\left(1+\frac{1}{k}\right)\right)=\sum _{k=1}^{N}{\left(-1\right)}^{k-1}\frac{1}{k}-\sum _{k=1}^{N}{\left(-1\right)}^{k-1}\mathrm{log}\left(1+\frac{1}{k}\right)$
$=\sum _{k=1}^{N}{\left(-1\right)}^{k-1}\frac{1}{k}-\mathrm{log}\left(\prod _{k=1}^{N}{\left(1+\frac{1}{k}\right)}^{{\left(-1\right)}^{k-1}}\right)$
giving
$\sum _{k=1}^{\mathrm{\infty }}{\left(-1\right)}^{k-1}\left(\frac{1}{k}-\mathrm{log}\left(1+\frac{1}{k}\right)\right)=\mathrm{log}2-\mathrm{log}\left(\frac{\pi }{2}\right)$
Finally we obtain Ulgelmorgs

Consider the series
$S=\sum _{k=1}^{\mathrm{\infty }}\left[\frac{1}{2k}-\mathrm{ln}\left(1+\frac{1}{2k}\right)\right]$
for which, by using the logarithm in series form, it becomes
$S=\sum _{k=1}^{\mathrm{\infty }}\left[\frac{1}{2k}-\frac{1}{2k}+\sum _{n=2}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n}}{n{\left(2k\right)}^{n}}\right]$
$=\sum _{k=1}^{\mathrm{\infty }}\sum _{n=2}^{\mathrm{\infty }}\frac{{\left(-\frac{1}{2}\right)}^{n}}{n{k}^{n}}$
$=\sum _{n=2}^{\mathrm{\infty }}\frac{{\left(-\frac{1}{2}\right)}^{n}}{n}\zeta \left(n\right)$
$=\sum _{n=1}^{\mathrm{\infty }}\frac{{\left(-\frac{1}{2}\right)}^{n+1}}{n+1}\zeta \left(n+1\right)$
Now the generating function for the zeta function is given by
$\sum _{n=1}^{\mathrm{\infty }}\zeta \left(n+1\right){\left(-1\right)}^{n+1}{x}^{n}=\gamma +\psi \left(x+1\right)$
$\sum _{n=1}^{\mathrm{\infty }}\frac{{\left(-x\right)}^{n+1}}{n+1}\zeta \left(n+1\right)=\gamma x+\mathrm{ln}\mathrm{\Gamma }\left(x+1\right)$
Letting $x=\frac{1}{2}$ in this series leads to the value of the series for S, namely
$\sum _{k=1}^{\mathrm{\infty }}\left[\frac{1}{2k}-\mathrm{ln}\left(1+\frac{1}{2k}\right)\right]=\frac{1}{2}\mathrm{ln}\left(\frac{\pi }{4}\right)+\frac{\gamma }{2}$

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