TITLE: Evaluate $\lim\limits_{n \to \infty} \frac 1{n\log \left ( 1+\frac1n \right ) }$
QUESTION [0 upvotes]: $$\lim\limits_{n \to \infty} \frac 1{n\log \left ( 1+\frac1n \right ) }$$
I start by writing  $\lim\limits_{n \to \infty} \frac 1n=0$, and  $\lim\limits_{n \to \infty} \frac 1{n+\log \left ( 1\frac1n \right ) }=0$
Since limit exists, by multiplication law, $\lim\limits_{n \to \infty} \frac 1{n\log \left ( 1+\frac1n \right ) }=0$
Hence the series $ \sum_{n}^{\infty }\frac{1}{n\log 1+\frac{1}{n}}$ converge.
However, the $\lim n \to \infty \frac{1}{n\log \left ( 1+\frac{1}{n} \right ) }$ is given to be 1 in the solution, would really like to know where I've done wrong.
Thanks for the help.

REPLY [0 votes]: We have that
$$ n+\log \left ( 1+\frac1n \right )=n+\frac1n\frac{\log \left ( 1+\frac1n \right )}{\frac1n} \to \infty+0\cdot 1=\infty$$
and therefore
$$\lim\limits_{n \to \infty} \frac 1{n+\log \left ( 1+\frac1n \right ) }=0$$
which is indeed correct.
For the other one we have
$$\lim\limits_{n \to \infty} \frac 1{n\log \left ( 1+\frac1n \right ) }=\lim\limits_{n \to \infty} \frac 1{\log \left ( 1+\frac1n \right )^n }=\frac1{\log e}=1$$
and therefore the series $\sum_{n}^{\infty }\frac{1}{n\log 1+\frac{1}{n}}$ diverges since the necessary condition $a_n \to 0$ is not satisfied.
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