Genesis Gibbs

2022-09-06

Series for logarithms
This is more of a challenge than a question, but I thought I'd share anyway. Prove the following identities, and prove that the pattern continues.
$\sum _{n=0}^{\mathrm{\infty }}\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)=\mathrm{ln}2$
$\sum _{n=0}^{\mathrm{\infty }}\left(\frac{1}{3n+1}+\frac{1}{3n+2}-\frac{2}{3n+3}\right)=\mathrm{ln}3$
$\sum _{n=0}^{\mathrm{\infty }}\left(\frac{1}{4n+1}+\frac{1}{4n+2}+\frac{1}{4n+3}-\frac{3}{4n+4}\right)=\mathrm{ln}4$
$\mathrm{e}\mathrm{t}\mathrm{c}.$

Dayana Powers

Let $m\ge 2$. Put:
${S}_{m}=\sum _{n\ge 0}\left(\frac{1}{mn+1}+\cdots +\frac{1}{mn+m-1}-\frac{m-1}{mn+m}\right)$
As $\frac{1}{mn+r}={\int }_{0}^{1}{x}^{mn+r-1}dx$, We have
${S}_{m}={\int }_{0}^{1}\frac{1+x+\cdots +{x}^{m-2}-\left(m-1\right){x}^{m-1}}{1-{x}^{m}}dx$
But
$1+x+\cdots +{x}^{m-2}-\left(m-1\right){x}^{m-1}=\left(1-{x}^{m-1}\right)+\cdots +\left({x}^{m-2}-{x}^{m-1}\right)$
Hence
$1+x+\cdots +{x}^{m-2}-\left(m-1\right){x}^{m-1}=\left[\left(1-x\right)\left(1+x+\cdots {x}^{m-2}\right)\right]+\cdots +\left[{x}^{m-2}\left(1-x\right)\right]$
and
$\left[\left(1+x+\cdots {x}^{m-2}\right)\right]+\cdots +\left[{x}^{m-2}\right]=1+2x+\cdots +\left(m-1\right){x}^{m-2}$
Thus
${S}_{m}={\int }_{0}^{1}\frac{\sum _{k=0}^{m-2}\left(k+1\right){x}^{k}}{1+x+\cdots +{x}^{m-1}}dx={\int }_{0}^{1}\frac{{P}^{\mathrm{\prime }}\left(x\right)}{P\left(x\right)}dx=\left[\mathrm{log}\left(1+x+\cdots +{x}^{m-1}\right){\right]}_{0}^{1}=\mathrm{log}\left(m\right)$

firmezas1

For any $m\in \mathbb{N},$
$\sum _{k=0}^{\mathrm{\infty }}\left(\frac{1}{mk+1}+\frac{1}{mk+2}+...+\frac{1}{mk+m-1}-\frac{m-1}{mk+m}\right)$
$=\underset{n\to \mathrm{\infty }}{lim}\left(\sum _{k=0}^{mn}\frac{1}{mk+1}+...+\frac{1}{mk+m-1}-\frac{m-1}{mk+m}\right)$
$=\underset{n\to \mathrm{\infty }}{lim}\left(\sum _{k=1}^{mn}\frac{1}{k}-m\sum _{k=1}^{n}\frac{1}{mk}\right)$
$=\underset{n\to \mathrm{\infty }}{lim}\left(\sum _{k=1}^{mn}\frac{1}{k}-\sum _{k=1}^{n}\frac{1}{k}\right)$
$=\underset{n\to \mathrm{\infty }}{lim}\left(\sum _{k=1}^{mn}\frac{1}{k}-\mathrm{ln}\left(mn\right)+\mathrm{ln}\left(mn\right)-\sum _{k=1}^{n}\frac{1}{k}\right)$
$=\underset{n\to \mathrm{\infty }}{lim}\left(\sum _{k=1}^{mn}\frac{1}{k}-\mathrm{ln}\left(mn\right)+\mathrm{ln}\left(n\right)-\sum _{k=1}^{n}\frac{1}{k}\right)+\mathrm{ln}\left(m\right)$
$=\gamma -\gamma +\mathrm{ln}\left(m\right)=\mathrm{ln}\left(m\right),$
where $\gamma$ is the Euler-Mascheroni constant.

Do you have a similar question?