amolent3u

## Answered question

2022-01-03

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

### Answer & Explanation

Philip Williams

Beginner2022-01-04Added 39 answers

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

Beginner2022-01-05Added 35 answers

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

Expert2022-01-11Added 669 answers

$\sum _{n=1}^{m}\mathrm{log}\left(\frac{n+1}{n}\right)=\sum _{n=1}^{m}\left(\mathrm{log}\left(n+1\right)-\mathrm{log}n\right)=\mathrm{log}\left(m+1\right)$
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.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?