Convergence of the series $\sum\limits_{n=1}^{\infty}\frac1n\log\left(1+\frac1n\right)$.

Question

I am trying to check the convergence or divergence of the series $\displaystyle\sum_{n=1}^{\infty}\dfrac1n\log\left(1+\dfrac1n\right)$.

My attempt: for a finite $p$,\begin{align}\displaystyle\sum_{k=n}^{n+p}\dfrac1k\log\left(1+\dfrac1k\right)&\lt\dfrac1n\displaystyle\sum_{k=n}^{n+p}\log\left(1+\dfrac1k\right)\\&=\dfrac1n\log\large\Pi_{k=n}^{n+p}\left(\dfrac{k+1}{k}\right)\\&=\dfrac1n\log\left(1+\dfrac{p+1}{n}\right)\\&\lt\dfrac1n\log2,\text{ for large $n$ and $p$ is finite.}\\&\lt\varepsilon\end{align} Hence the series converges.

Answer 1

Because $$\frac{\frac{1}{n}\ln(1+\frac{1}{n})}{\frac{1}{n^2}}\rightarrow1$$ and $$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}.$$

Answer 2

$a_n:= (1/n)\log (1+1/n);$

Recall : $\lim_{n \rightarrow \infty}(1+1/n)^n=e$.

Hence $(1+1/n)^n$ is bounded by a $M$, real, positive.

Then $(1+1/n)^n <M$, and with

$\log (1+1/n)^n < \log M <M$:

$a_n = (1/n^2) \log(1+1/n)^n <M/n^2$.

Comparison test: $M \sum 1/n^2$ converges.

Answer 3

Your argument is not correct. For the Cauchy criterion you have to show that for every $\varepsilon > 0$ there is an $N \in \Bbb N$ such that $$ \sum_{k=n}^{n+p}\dfrac1k\log\left(1+\dfrac1k\right) < \varepsilon $$ for all $n \ge N$ and all $p \ge 0$. So you can not “fix” $p$ and assume that $$ \dfrac1n\log\left(1+\dfrac{p+1}{n}\right)\lt\dfrac1n\log2 \, . $$

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But using the “well-known” estimate $\log x \le x-1$ one gets $$ 0 \le \frac1n\log\left(1+\dfrac1n\right) \le \frac{1}{n^2} $$ and that implies the convergence by the “squeeze theorem.”