For which $p$ does $\sum\limits_{n=1}^{\infty}\left(1-\frac{p \ln(n)}{n}\right)^{n}$ converge? $$\sum\limits_{n=1}^{\infty}\left(1-\frac{p \ln(n)}{n}\right)^{n}$$
For which $p$ does it converge?
I tried to write it in a different way :
$$\left(1-\frac{p \ln(n)}{n}\right)^{n} = \left[\left(1-\frac{1}{\frac{n}{p\ln(n)}}\right)^{\frac{n}{p\ln(n)}}\right]^{-p\ln(n)}  \sim e^{-p\ln(n)}$$
so i have to compare the original one with $e^{-p\ln(n)}$ which is convergent for $p>1$.
I am not sure if that is correct and how to prove that the lim exist with the comparison test.
 A: By the Taylor series expansion of $x \mapsto \ln(1-x)$, $0\le x<1$, near $0$, one has
$$
\ln(1-x)=-x+O(x^2).
$$
Then one may write, as $n \to \infty$,
\begin{align}
\left(1-\frac{p \ln(n)}{n}\right)^{n}&=e^{n\ln \left(1-\frac{p \ln(n)}{n} \right)}
\\\\&=e^{-n\frac{p \ln(n)}{n} +n\,O\left(\frac{\ln^2 n}{n^2}\right)}
\\\\&=e^{\ln \frac{1}{n^p}}\times e^{O\left(\frac{\ln^2 n}n\right)}
\\\\&=\frac{1}{n^p}\times e^{o(1)}
\\\\&\sim \frac{1}{n^p}
\end{align}
the given series converges iff $p>1$ by the limit comparison test.
A: Both the OP's suggestion and Olivier Oloa's answer use the limit comparison test.
Actually, it's not that simple. The limit comparison test requires 2 sequences that are either both always positive, or both always negative. 
However, if $p>e$, $\left(1-\frac{p\ln(n)}{n}\right)^n$ is not always positive.
Now, one can show pretty easily that for every $p$, $\exists n_p \in \mathbb{N}$ such that $\forall n > n_p, \left(1-\frac{p\ln(n)}{n}\right)^n>0$ (that is to say, that although this sequence is not always positive, there is a certain point from where it's always positive).
And then we can apply the limit comparison test to sequence new sequence optained by starting at this $n_p$ instead of $n=1$. And since this is a limit test (asymptotic by nature), an offset of $n_p$ doesn't change the result.
