How find the $\sum_{n=1}^{\infty}\frac{1}{n^p}\left(1-\frac{x\ln{n}}{n}\right)^n$ convergent Question:
Study the series
$$\sum_{n=1}^{\infty}\dfrac{1}{n^p}\left(1-\dfrac{x\ln{n}}{n}\right)^n$$
convergence,when $p$ and $x$ such what conditions?
My try:
since
$$\dfrac{1}{n^p}\left(1-\dfrac{x\ln{n}}{n}\right)^n=\dfrac{1}{n^p}e^{n\ln{\left(1-\dfrac{x\ln{n}}{n}\right)}}=\dfrac{1}{n^p}e^{n\left(-\dfrac{xln{n}}{n}+o(-\dfrac{xln{n}}{n})\right)}=\dfrac{1}{n^p}e^{-x\ln{n}+o(-x\ln{n})}$$
and my try is usefull? and How solve this problem, Thank you.

We can find easily that
$$\dfrac{1}{n^p}\left(1-\dfrac{x\ln{n}}{n}\right)^n\sim_\infty\dfrac{1}{n^p}\dfrac{1}{n^x}=\dfrac{1}{n^{p+x}}$$
so the necessary and sufficient condition of the convergence is
$$p+x>1$$
Added Following the Sundaycat's request 
Recall that $\ln(1- x)\sim_0 -x$ so
$$\left(1-\dfrac{x\ln{n}}{n}\right)^n=\exp\left(n\ln\left(1-\dfrac{x\ln{n}}{n}\right)\right)\sim_\infty\exp(-x\ln n)=\frac{1}{n^x}$$
