Ways of showing \sum_{n=1}^\infty\ln(1+\frac{1}{n}) to be divergent

amolent3u

amolent3u

Answered question

2022-01-03

Ways of showing n=1ln(1+1n) to be divergent

Answer & Explanation

Philip Williams

Philip Williams

Beginner2022-01-04Added 39 answers

Notice the following:
log(1+1n)=log(n+1n)=log(n+1)log(n)
Hence
k=1nlog(1+1k)=log(n+1)
Corgnatiui

Corgnatiui

Beginner2022-01-05Added 35 answers

This is a special case of: Suppose f(1)=0,f(1)>0. Then f(1+1n)=
Proof: From the definition of the derivative (no Taylor necessary), we have
f(1+h)f(1)h=f(1+h)h>f(1)2
for small h>0. Thus
f(1+1n)>f(1)21n
for large n. By the comparison test we're done.
Vasquez

Vasquez

Expert2022-01-11Added 669 answers

n=1mlog(n+1n)=n=1m(log(n+1)logn)=log(m+1)
The partial sums clearly diverge. Alternatively using the cauchy condensation test the series converges iff the series
n=12nln(1+12n)
converges. The transformed series diverges since the terms don't go to zero and so the original series diverges.

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