Check the convergence of the series $ \sum\left (1+\frac{1}{\sqrt n}\right)^{-n^{3/2}}$ 
Check the convergence of the series $$ \sum_{n=1}^{\infty} \left(1+\frac{1}{\sqrt n}\right)^{-n^{3/2}}$$

I checked this equation with the test for divergence but it came out 0. So, now I don't know the nature of the following equation. Root test and ratio test and limit test are not going to work (or maybe I am doing it wrong) and I don't know how to use comparison test in this question.
Besides, I am having problem with the power i.e $-n^{\frac{3}{2}}$ in the following question, how to remove it?
 A: To check the convergence of the series, one can forget the root test or equivalents or whatnots, and work directly with "hard" inequalities, since, for every $x$ in $[0,1]$, $$1+x\geqslant2^x\tag{$\ast$}$$ hence, for every $n\geqslant1$, $$\left(1+\frac1{\sqrt{n}}\right)^{-n\sqrt{n}}\leqslant\left(2^{1/\sqrt{n}}\right)^{-n\sqrt{n}}=2^{-n}$$ Thus, the series converges (and its sum is less than $1$).

To show $(\ast)$, note that the function $u:x\mapsto2^x$ is convex, that $y=1+x$ is the equation of the chord of the graph of $u$ between the points of affixes $x=0$ and $x=1$, and that $(\ast)$ simply states that the graph of $u$ is below its chord on $[0,1]$.
A: Apply the root test, then
$$\lim_{n\to\infty} \sqrt[n]{|a_n|}=\lim_{n\to\infty}\left(1+\frac{1}{\sqrt n}\right)^{-\frac{n^{\frac{3}{2}}}{n}}=\lim_{n\to\infty}\left(1+\frac{1}{\sqrt n}\right)^{-\sqrt{n}}=e^{-1}$$
where we used the known limit
$$\lim_{x\to +\infty}\left(1+\frac{1}{x}\right)^x=e.$$
Since the limit $e^{-1}$ is less than $1$, the series is convergent.
A: The $n$th term is
$$\tag 1 \left (\frac{1}{(1+1/\sqrt n)^{\sqrt n}}\right)^n.$$
Now $(1+1/\sqrt n)^{\sqrt n} \to e >2,$ hence $(1)\le (1/2)^n$ for large $n.$ Thus the series converges by the comparison test.
A: By limit comparison test
and using the fact that when  $n\to +\infty $,
$$-n^\frac 32\ln (1+\frac {1}{\sqrt {n}})=$$
$$-n^\frac 32(\frac {1}{\sqrt {n}}-\frac {1}{2n}+\frac{1}{3n\sqrt {n}}-\frac {1}{4n^2}(1+\epsilon (n)) $$
thus
$$u_n\sim e^{-n+\frac {\sqrt {n}}{2}-\frac {1}{3}+\frac {1}{4\sqrt {n}}    }$$
the series converges.
