For which $p$ does $\sum\limits_{n=1}^{\infty}\left(1-\frac{p \ln(n)}{n}\right)^{n}$ converge?

Question

$$\sum\limits_{n=1}^{\infty}\left(1-\frac{p \ln(n)}{n}\right)^{n}$$

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.

Answer 1

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.

Answer 2

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.