Series convergence: bound needed Why should the following series converge? Can't find a good bound
$$\sum_{n=1}^\infty{\exp{\left( -\frac{n^{1-\alpha}}{\sqrt{\log{n}}}\right)}}$$ for $\alpha<1$
 A: By L'Hospital's rule, it is easily seen that
$$
2\log^{3/2}{n}=o(n^{\epsilon})
$$
where $\epsilon>0$. So for large $n>N$
$$
2\log{n}<\frac{n^{\epsilon}}{\log^{1/2}{n}}
$$
And 
$$
\exp\left(-\frac{n^{\epsilon}}{\log^{1/2}{n}}\right)<\exp\left(-2\log{n}\right)=\frac1{n^2}
$$
So 
$$
\sum_{n=N}^{\infty}\exp\left(-\frac{n^{\epsilon}}{\log^{1/2}{n}}\right)<\sum_{n=N}^{\infty}\frac1{n^2}<\infty
$$
A: Delete the term with $n=0$ because it contains $1/ \log 1 =1/0$.... We have  $\lim _{m \to \infty}  ( \log m )/m=0$, so for fixed  $b > 0$  we have $0 < (\log m)/m < 2b$  for all sufficiently large $m$, so  for all sufficiently large $n$ we have $0 < (\log n^{2b})/n^{2b} < 2b$, equivalently, $0 < \sqrt {\log n} < n^b$.  So let $0 < b < 1- \alpha$ and let $c=1- \alpha -b$. ( Note $c > 0$.)   For sufficiently large $n$ we have $$n^{1-\alpha}/\sqrt {\log n} > n^c $$ so $$0 < \text{exp}(-n^{1-\alpha}/\sqrt {\log n}) <  \text{exp}(-n^c).$$ Now the function $f(n)= \exp {(-n^c)}$ for $n \in N$ is decreasing and positive and so is the continuous function  $g(x)=\exp {(-x^c)}$ for $x  > 1$, while $f(n)=g(n)$ for $n \in N$.Thus we may use the integral test: If $$J=\int_{x=2}^\infty g(x)dx < \infty$$ then the summation  $\sum_{n=2}^\infty \exp(-n^c)$ is finite.Make a change of variable: $x=y^c$, so $$J= \int_{y=d}^\infty (cy^{1-1/c}\exp y)^{-1}  dy$$ where $d=2^{1/c} > 0$ .Now choose $k \in N$ with $k+1-1/c > 2$. We have $\exp y > y^k/k!$ for all positive $y$ so $$J <
\int_d^\infty k!/(cy^2) dy < \infty$$ so the series converges.
