amolent3u

2022-01-03

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

Philip Williams

Beginner2022-01-04Added 39 answers

Notice the following:

$\mathrm{log}(1+\frac{1}{n})=\mathrm{log}\left(\frac{n+1}{n}\right)=\mathrm{log}(n+1)-\mathrm{log}\left(n\right)$

Hence

$\sum _{k=1}^{n}\mathrm{log}(1+\frac{1}{k})=\mathrm{log}(n+1)\to \mathrm{\infty}$

Hence

Corgnatiui

Beginner2022-01-05Added 35 answers

This is a special case of: Suppose $f\left(1\right)=0,{f}^{\prime}\left(1\right)>0$ . Then $\sum f(1+\frac{1}{n})=\mathrm{\infty}$

Proof: From the definition of the derivative (no Taylor necessary), we have

$\frac{f(1+h)-f\left(1\right)}{h}=\frac{f(1+h)}{h}>\frac{{f}^{\prime}\left(1\right)}{2}$

for small$h>0$ . Thus

$f(1+\frac{1}{n})>\frac{{f}^{\prime}\left(1\right)}{2}\cdot \frac{1}{n}$

for large$n$ . By the comparison test we're done.

Proof: From the definition of the derivative (no Taylor necessary), we have

for small

for large

Vasquez

Expert2022-01-11Added 669 answers

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