Find $\lim\limits_{n \to +\infty}\frac {n}{(\log(n+1))^n}$ I'm studying the convergence of the series
$$\sum_{n=1}^\infty \frac {n}{(\log(n+1))^n}$$

*

*$\frac {n}{(\log(n+1))^n}>0, \forall n \ge 1$


*throught the root test I can proof the series converges but I can't solve the limit
$\lim\limits_{n \to +\infty}\frac {n}{(\log(n+1))^n}$ should be $0$ .
 A: I think ratio test might be easier 
$$\color{red}{0\leq}\lim\limits_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim\limits_{n\rightarrow\infty} \frac{\frac {n+1}{(\log(n+2))^{n+1}}}{\frac {n}{(\log(n+1))^n}}=\\
\lim\limits_{n\rightarrow\infty}\frac{n+1}{n}\left(\frac{\log(n+1)}{\log(n+2)}\right)^n\frac{1}{\log(n+2)}\leq ...$$
since $\log(n+1)< \log(n+2)$ thus $\left(\frac{\log(n+1)}{\log(n+2)}\right)^n<1$
$$...\leq\lim\limits_{n\rightarrow\infty}\frac{n+1}{n}\frac{1}{\log(n+2)}=
\lim\limits_{n\rightarrow\infty}\frac{n+1}{n}\cdot \lim\limits_{n\rightarrow\infty}\frac{1}{\log(n+2)}=\\
1\cdot \lim\limits_{n\rightarrow\infty}\frac{1}{\log(n+2)}
\color{red}{=0}$$
Thus the series converges and the $\lim\limits_{n \to \infty}\frac {n}{(\log(n+1))^n}=0$
A: Using the root test
$$ \sqrt[1/n]{n/\ln(n+1)^n} = n^{1/n}/\ln(n+1). $$
For the guy above we have
$$ n^{1/n} = e^{\ln(n^{1/n})} = e^{\ln(n)/n}, $$
so that we can study the limit
$$ \lim_n \ln(n)/n \overset{Hopital}{=} \lim_n 1/n=0. $$
Combining all together, along with the continuity of $e^x$, we get
$$ \limsup_n \sqrt[1/n]{n/\ln(n+1)^n} = \limsup_n e^{\ln(n)/n}/\ln(n+1) =0.$$
Note that a full argument would be "consider $f:\mathbb{R}^+\to \mathbb{R},x\mapsto \sqrt[1/x]{x/\ln(x+1)^n}$ etc.", you find that the limsup exists, so that the lim exists and it is $0$ being the function positive (by comparison). Therefore for any sequence $\{x_n\}_{n\in\mathbb{N}}\subset \mathbb{R},x_n \to \infty$, you have $f(x_n)\to 0$. Pick $x_n=n$ and you are done.
A: Let
$$a_n= \frac {n}{(\log(n+1))^n}\to0$$
indeed by ratio test
$$\frac{a_{n+1}}{a_n}=\frac {n+1}{ (\log(n+2))^{n+1} } \frac {(\log(n+1))^n }{ n }=\frac{n+1}{n} \frac{1}{\log(n+2)} \left(\frac { \log(n+1) }{  \log(n+2) }\right)^n\to0$$
indeed
$$\frac{n+1}{n}\to1, \quad  \frac{1}{\log(n+2)}\to 0, \quad 0< \left(\frac { \log(n+1) }{  \log(n+2) }\right)^n <1$$
