On the nature of a series involving ratios and power of logarithm Is the series $$\sum_{n=1}^\infty \left(\dfrac {\log n}{\log (n+1)} \right)^{n^2\log n}$$ convergent? I tried to apply some test, or compare with known series, but with no success; it looks too complicated. Please help. Thanks in advance.
 A: From a Taylor expansion to low-order terms:
$$\begin{align}
\frac{\log n}{\log(n+1)}
&= \frac{\log n}{\log n+\log(1+\frac{1}{n})}
= \frac{1}{1+\frac{1}{n\log n}+o\left(\frac{1}{n\log n}\right)}
\\&= 1-\frac{1}{n\log n}+o\left(\frac{1}{n\log n}\right)
\end{align}$$
from which
$$\begin{align}
\left(\frac{\log n}{\log(n+1)}\right)^{n^2\log n}
&=\exp\left(n^2\log n\log\left(\frac{\log n}{\log(n+1)}\right)\right)
\\
&=\exp\left(n^2\log n\log\left(1-\frac{1}{n\log n}+o\left(\frac{1}{n\log n}\right)\right)\right)\\
&=\exp\left(n^2\log n\left(-\frac{1}{n\log n}+o\left(\frac{1}{n\log n}\right)\right)\right)\\
&=\exp\left(-n+o\left(n\right)\right)
=\frac{1}{e^{n+o(n)}}
\end{align}$$
Thus, by comparison, the series will converge, since $\sum_n \frac{1}{e^n}$ does.
A: Hint. Note that
\begin{align*}
\left(\dfrac {\log n}{\log (n+1)} \right)^{n^2\log n}&=\left(1+\frac{\log(1+1/n)}{\log(n)}\right)^{-n^2\log n}\\&=
\exp\left(-n^2\log(n)\log\left(1+\frac{\log(1+1/n)}{\log(n)}\right)\right)\\
&=
\exp\left(-n\cdot n\log(n)\cdot\log\left(1+\frac{1+o(1)}{n\log(n)}\right)\right)\\
&=
\exp\left(-n(1+o(1)\right).
\end{align*}
