How to show the convergence of the following series: $$\sum\limits_{n=2}^{\infty} \,\left(1-\frac{1}{(\ln\,n)^{k}}\right)^n,$$ where $k$ is a fixed positive integer. 
I have tried the standard tests by the are all inconclusive.
Many thanks for your help. 
 A: We will show below that $\displaystyle\lim_{x\to\infty}x^2\left(1-\frac{1}{(\ln x)^k}\right)^x=0$, 
so that $\displaystyle\lim_{n\to\infty}\left(1-\frac{1}{(\ln n)^k}\right)^n\div\left(\frac{1}{n^2}\right)=0$
and therefore the series converges using the Limit Comparison Test with $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n^2}$:

$\hspace{.3 in}\displaystyle\lim_{t\to\infty}\left(\frac{e^t}{t^{k+1}}\right)=\lim_{t\to\infty}\frac{e^t}{(k+1)!}=\infty$ using L'Hospital's Rule k+1 times; so
$\hspace{.3 in} \displaystyle\lim_{t\to\infty}\left(\frac{e^t}{t^k}-2t\right)=\lim_{t\to\infty}t\left(\frac{e^t}{t^{k+1}}-2\right)=\infty$. $\;\;$Then
$\hspace{.3 in}\displaystyle\lim_{t\to\infty}\big[e^t\left(\ln(t^k)-\ln(t^k-1)\right)-2t\big]=\infty\;\;$ since $\ln(t^k)-\ln(t^k-1)>\frac{1}{t^k}$ and therefore
$\hspace{.3 in}\displaystyle\lim_{t\to\infty}\big[2t+e^t\left(\ln(t^k-1)-\ln(t^k)\right)\big]=-\infty$. $\hspace{.2 in}$Letting $t=\ln x$ gives
$\hspace{.3 in}\displaystyle\lim_{x\to\infty}\left[2\ln x+x\left(\ln((\ln x)^k-1)-\ln(\ln x)^k\right)\right]=-\infty$, so
$\hspace{.3 in}\displaystyle\lim_{x\to\infty}\ln\left[x^2\left(1-\frac{1}{(\ln x)^k}\right)^x\right]=-\infty$ and thus $\displaystyle\lim_{x\to\infty}x^2\left(1-\frac{1}{(\ln x)^k}\right)^x=0$.
A: $$\sum\limits_{n=2}^{\infty} \,\left(1-\frac{1}{(\ln\,n)^{k}}\right)^n=\sum\limits_{n=2}^{\infty} \,\left(1-\frac{1}{(\ln\,n)^{k}}\right)^{(\ln\,n)^{k}\cdot\frac{n}{(\ln\,n)^{k}}}\leqslant$$ $$c_1+\sum\limits_{n=2}^{\infty} e^{-\sqrt{n}}\leqslant c_2+\sum\limits_{t=1}^{\infty} t^2e^{-t}\leqslant c_3+\sum\limits_{t=1}^{\infty} e^{-0.9t}<\infty,$$with some $c_i\in\mathbb{R}$.
