 amolent3u

2022-01-03

Ways of showing $\sum _{n=1}^{\mathrm{\infty }}\mathrm{ln}\left(1+\frac{1}{n}\right)$ to be divergent Philip Williams

Notice the following:
$\mathrm{log}\left(1+\frac{1}{n}\right)=\mathrm{log}\left(\frac{n+1}{n}\right)=\mathrm{log}\left(n+1\right)-\mathrm{log}\left(n\right)$
Hence
$\sum _{k=1}^{n}\mathrm{log}\left(1+\frac{1}{k}\right)=\mathrm{log}\left(n+1\right)\to \mathrm{\infty }$ Corgnatiui

This is a special case of: Suppose $f\left(1\right)=0,{f}^{\prime }\left(1\right)>0$. Then $\sum f\left(1+\frac{1}{n}\right)=\mathrm{\infty }$
Proof: From the definition of the derivative (no Taylor necessary), we have
$\frac{f\left(1+h\right)-f\left(1\right)}{h}=\frac{f\left(1+h\right)}{h}>\frac{{f}^{\prime }\left(1\right)}{2}$
for small $h>0$. Thus
$f\left(1+\frac{1}{n}\right)>\frac{{f}^{\prime }\left(1\right)}{2}\cdot \frac{1}{n}$
for large $n$. By the comparison test we're done. Vasquez

$\sum _{n=1}^{m}\mathrm{log}\left(\frac{n+1}{n}\right)=\sum _{n=1}^{m}\left(\mathrm{log}\left(n+1\right)-\mathrm{log}n\right)=\mathrm{log}\left(m+1\right)$
The partial sums clearly diverge. Alternatively using the cauchy condensation test the series converges iff the series
$\sum _{n=1}^{\mathrm{\infty }}{2}^{n}\mathrm{ln}\left(1+\frac{1}{{2}^{n}}\right)$
converges. The transformed series diverges since the terms don't go to zero and so the original series diverges.

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