Prove that $\sum_{n=1}^{\infty}\exp\left(-n^{\varepsilon}\right)$ converges. I have the problem to prove, that the following series converges:
\begin{equation}
\sum_{n=1}^{\infty}\exp\left(-n^{\varepsilon}\right),
\end{equation}
where $\varepsilon > 0$.
I tried everything. I tried the ratio and root test, but I don't come to a solution. Can someone help me?
Thanks!
 A: Because $\lim_{n\to\infty}\frac{n^2}{\exp(n^{\epsilon})}=0$ for any $\epsilon>0$ by twice L'Hospitals rule, you can infer that for a certain $N$ we have $n^2<\exp(n^{\epsilon})$ for all $n>N$, and from there the comparison test gives convergence.
A: For $\epsilon>0$, find an $N$ such that $N\epsilon>1$. Now use Taylor formula to obtain that $e^{u}>Cu^{N}$, $u>0$ for some constant $C>0$ that depending only on $N$. Then $e^{-n^{\epsilon}}<c\dfrac{1}{n^{N\epsilon}}$ and we have $\displaystyle\sum\dfrac{1}{n^{N\epsilon}}<\infty$ by $p$-series.
A: Observe that $$\lim_{n \to \infty} \frac{2 \log n}{n^\epsilon} = 0$$ for every $\epsilon > 0$.  In particular, there exists an $N$ with the property that 
$$n \ge N \implies 2 \log n < n^\epsilon \implies e^{-n^\epsilon} < e^{-2 \log n} \implies e^{-n^\epsilon} < \frac 1{n^2}.$$
Now use the comparison test.
A: Using the Cauchy condensation test $\sum_{n=1}^{\infty}\exp\left(-n^{\varepsilon}\right)$ is convergent iff
$$
\sum_{n=1}^\infty\frac{2^n}{e^{2^{n\epsilon}}} =
\sum_{n=1}^\infty\frac1{e^{2^{n\epsilon} - n\log 2}}
$$
is convergent, but as $2^{n\epsilon} - n\log 2 > n$ for n large enough, 
$$\frac1{e^{2^{n\epsilon} - n\log 2}}< \frac1{e^n}$$
for $n$ large enough.
