Convergence of the series $\sum\limits_n \left(1-\frac{K}{n^{1-\epsilon}\sqrt{\log n}} \right)^n$ for $\epsilon>0$ This is not a homework problem. It has come up in my research. I am trying to show that $$\sum_n \left(1-\frac{K}{n^{1-\epsilon}\sqrt{\log n}} \right)^n$$ converges for  $\epsilon&gt;0$. I have no idea how to approach the problem.
Start by an asymptotic expansion of $\log\left(1-\frac{K}{n^{1-\epsilon}\sqrt{\log n}}\right)$, that is 
$$\log\left(1-\frac{K}{n^{1-\epsilon}\sqrt{\log n}}\right)=-\frac K{n^{1-\varepsilon}\sqrt{\log n}}-\frac{K^2}{2n^{2-2\varepsilon}\log n}(1+\delta_n),$$
where $\lim_{n\to\infty}\delta_n=0$. This gives 
$$\exp\left(n\log\left(1-\frac{K}{n^{1-\epsilon}\sqrt{\log n}}\right)\right)=
\exp\left(-\frac{n^\varepsilon}{\sqrt{\log n}}-\frac{K^2}{2n^{1-2\varepsilon}\log n}(1+\delta_n)\right).$$
The main term is of order
$$a_n=\left(1-\frac{1}{n^{1-\epsilon}\sqrt{\ln n}}\right)^n\sim e^{-\dfrac{n^\varepsilon}{\sqrt{\ln n}}}\leq e^{-n^\varepsilon}\leq \frac{1}{n^2}$$
for arbitrary large $n$, so the series converges.