To test the convergence of a series whose nth term is $a_{n}=\left({\frac{\log n}{\log(n+1)}}\right)^{n^2\log n}$ using Root test Test the convergence of the series whose nth term is given using Root test
$$a_{n}=\left({\frac{\log n}{\log(n+1)}}\right)^{n^2\log n}$$
I know that if $\lim_{n \to \infty}(a_{n})^{1/n}\lt 1$,the series converges
So,
let
$$ \lim_{n \to \infty}\left({\frac{\log n}{\log(n+1)}}\right)^{n\log n}=y$$
Taking log on both sides,
$$\lim_{n \to \infty}n\log n\log\left({\frac{\log n}{\log(n+1)}}\right)=\log y$$
But this is leading to complicated expressions.Is there a simpler way to approach this?
 A: Hint
When $n$ is large
$$\frac{\log (n)}{\log (n+1)}\sim 1-\frac 1 {n \log(n)}$$
$$\log \left(\frac{\log (n)}{\log (n+1)}\right)\sim -\frac 1 {n \log(n)}$$
A: Note that
$$\left(\frac{\log n}{\log(n+1)}\right)^{n\log n}=\left(\frac{\log n}{\log n+\log(1+\frac1n)}\right)^{n\log n}=\left(\frac{\log n^n}{\log n^n+\log(1+\frac1n)^n}\right)^{\log n^n}=\left(\frac{1}{1+\frac{\log(1+\frac1n)^n}{\log n^n}}\right)^{\log n^n}\to\frac1e$$
indeed
$$\left(1+\frac{\log(1+\frac1n)^n}{\log n^n}\right)^{\log n^n}=e^{\log n^n\cdot \log \left(1+\frac{\log(1+\frac1n)^n}{\log n^n}\right)}=e^{\log n^n\cdot \left(\frac{\log(1+\frac1n)^n}{\log n^n}+o\left(\frac{1}{logn^n}\right)\right)}=e^{\left(\log(1+\frac1n)^n+o(1)\right)}\to e$$
A: I think you can construct an inequality first to simplify the expression:  
Because $0<\frac{log(n)}{log(n+1)}<1 \quad$ and  $\quad n^2\log(n)\ge n >1$ $\qquad$ $\forall n\in\mathbb N, n>1$
We have $$a_{n}=\left({\frac{\log n}{log(n+1)}}\right)^{n^2\log n}<\left({\frac{\log n}{log(n+1)}}\right)^{n^2}=b_{n}$$
We now apply root test to $\sum b_{n}$ and evaluate:
$$ lim_{n \rightarrow \infty}\left({\frac{\log n}{log(n+1)}}\right)^{n}=1$$
We get an indeterminate result. But, we can use Raabe's test as an extension:
$$ lim_{n \rightarrow \infty}\left(n\cdot\bigr({\frac{b_{n}}{b_{n+1}}}\bigr)^{n}-1\right)=\infty$$
This means $\sum b_{n}$ converges. Thus, the original series $\sum a_{n}$ converges.
