I am trying to prove whether \begin{equation*} \sum\limits_{n=1}^{\infty}\log(1 + \frac{1}n) \end{equation*} converges or diverges, but none of the normal tests (nth test, p test, etc. ) seem to work.
I was just wondering if someone wouldnt mind giving me a suggestion as to how i would show if this converges or diverges?
Thanks in advance C :)
$$\sum_{n=1}^{\infty}{\log(1 + \frac{1}{n})} = \sum_{n=1}^{\infty}{\log(\frac{n + 1}{n})} = \sum_{n=1}^{\infty}{\left(\log(n + 1) - \log(n)\right)} = \lim_{N \rightarrow \infty}{\left(\sum_{n=1}^{N}{{\left(\log(n + 1) - \log(n)\right)}}\right)} = \lim_{N \rightarrow \infty}{\log(N + 1)} \rightarrow \infty$$
Therefore,
$$\sum_{n=1}^{\infty}{\log(1 + \frac{1}{n})}$$ diverges.
for positive sequences, if $a_n \sim b_n$, then $\sum a_n $ converges $\iff \sum b_n$ converges
Since $\log(1 + \frac 1n) \sim \frac 1n$ and the latter diverges, we conclude that
$$\sum \log\left(1 + \frac 1n\right)$$ diverges, too
$\sum \ln(1+\frac{1}{n})=\sum \ln(\frac{n+1}{n})=\ln\Pi_{n=1}^\infty \frac{n+1}{n}=\ln(1\frac21\frac32...\frac{n+1}{n}\frac{n+2}{n+1}...)=\lim_{n\to \infty}\ln(n)=\infty$
It is well-known that the harmonic sequence $H_N=\sum_{k=1}^N \frac1k$ diverges to infinity. And, for $N\in\Bbb N$,
$$\sum_{n=1}^N\ln\left(1+\frac1n\right)=\sum_{k=1}^N\int_1^{1+\frac1n}\frac{dx}x\ge\sum_{k=1}^N\frac1n\cdot\frac{n}{n+1}=H_N-1$$
