Showing $\sum^\infty_{n=1} \frac 1 {e^{\sqrt n}} $ converges 
Show that $\displaystyle\sum^\infty_{n=1} \frac 1 {e^{\sqrt n}}$ converges.

My attempt: 
Using limit ratio test: $\displaystyle\lim_{n\to\infty} \frac {e^{\sqrt n}} {e^{\sqrt {n+1}}}=\lim_{n\to\infty}\frac {e^{\frac n 2}} {e^{\frac {n+1} 2}}=\lim_{n\to\infty}\frac {e^{\frac n 2}} {e^{\frac n 2+\frac 1 2}}=\lim_{n\to\infty}\frac 1 {e^{\frac n 2+\frac 1 2-\frac n 2}}=e^{-0.5}$
And since $e^{-0.5}<1$ the series converges.
Is that alright?
 A: The condensation test
works here:
if $f(n)$
is positive and decreasing,
then
$\sum f(n)$
converges if and only if
$\sum 2^n f(2^n)$
converges.
If 
$f(n) = e^{-\sqrt{n}}$,
then
$2^n f(n) 
= 2^n e^{-\sqrt{2^n}}
= e^{n \ln 2 -2^{n/2}}
$.
Since
$2^n > n^4$
for $n \ge 17$
(see a proof 
in my answer here:
Prove that $n^k < 2^n$ for all large enough $n$),
$ 2^{n/2}
> n^2
$
so
$e^{n \ln 2 -2^{n/2}}
<e^{n \ln 2 -n^2}
<e^{-n(n- \ln 2)}
$
and the sum of these
clearly converges.
Note:
The result I proved there is this:
If $n$ and $k$ are integers and $k≥2$ and $n≥k^2+1$, then $2^n>n^k$.
A: $$
\begin{split}
\displaystyle\sum^\infty_{n=1} \frac 1 {e^{\sqrt n}}&=\ 
\underbrace{e^{-\sqrt1}+e^{-\sqrt2}+e^{-\sqrt3}}_{3\text{ terms}}
+
\underbrace{e^{-\sqrt4}+e^{-\sqrt5}+e^{-\sqrt6}+e^{-\sqrt7}+e^{-\sqrt8}}_{5\text{ terms}}
+
\underbrace{e^{-\sqrt9}+\cdots + e^{-\sqrt{15}}}_{7\text{ terms}}
+
\cdots\\
&<\ 3\cdot e^{-\sqrt1}+5\cdot e^{-\sqrt4}+7\cdot e^{-\sqrt9}+\cdots+(2k+1)\cdot e^{-\sqrt{k^2}}+\cdots\\
&=\ 3\cdot e^{-1}+5\cdot e^{-2}+7\cdot e^{-3}+\cdots+(2k+1)\cdot e^{-k}+\cdots\\
&<\ 4\cdot e^{-1}+8\cdot e^{-2}+16\cdot e^{-3}+\cdots2^{k+1}\cdot e^{-k}+\cdots\\
&=\ \frac{4\cdot e^{-1}}{1-\frac{2}{e}}<\infty.
\end{split}
$$
A: The step

$$\lim_{n\to\infty} \frac {e^{\sqrt n}} {e^{\sqrt {n+1}}}=\lim_{n\to\infty}\frac {e^{\frac n 2}} {e^{\frac {n+1} 2}}$$

desperately needs justification. 
Since $$\frac {e^{\sqrt n}} {e^{\sqrt {n+1}}} = e^{\sqrt{n} - \sqrt{n+1}}$$ and $$\lim_{n \to \infty} \left( \sqrt{n} - \sqrt{n+1} \right) = \lim_{n \to \infty} \frac{-1}{\sqrt{n} + \sqrt{n+1}} = 0$$
you actually have $$\lim_{n \to \infty} \frac{e^{\sqrt n}} {e^{\sqrt {n+1}}} = 1.$$
A: The correct application of the ratio test would be as follows:
$$
\begin{align*}
\lim_{n\to\infty} \frac {e^{\sqrt n}} {e^{\sqrt {n+1}}}&=
\lim_{n \to \infty} e^{\sqrt n - \sqrt{n+1}} 
\\ & =
\exp\left[ \lim_{n \to \infty} \sqrt n - \sqrt{n + 1} \right]
\\ & =
\exp \left[ \lim_{n \to \infty} \frac{-1}{\sqrt n + \sqrt{n + 1}} \right] = 1
\end{align*}
$$
So, the ratio test fails.  In fact, we could use the comparison test with, for example, $\sum 1/n^2$.  It suffices to note that
$$
\lim_{n \to \infty} \frac{1/e^{\sqrt n}}{1/n^2} = 0
$$
To prove the above, we have
$$
\lim_{n \to \infty} \frac{1/e^{\sqrt n}}{1/n^2} = 
\lim_{n \to \infty} \frac{n^2}{e^{\sqrt n}} \overset{u = \sqrt n}{=} 
\lim_{u \to \infty} \frac{(u^2)^2}{e^u} = 
\lim_{u \to \infty} \frac{u^4}{e^u}
$$
from there, apply L'Hospital's rule.
A: A very simple answer
just occurred to me.
For
$\sum^\infty_{n=1} \frac 1 {e^{\sqrt n}}
$,
by looking at the
first few terms of the
power series,
$e^{\sqrt n}
\gt 1+\sqrt n
+\frac{(\sqrt n)^2}{2}
+\frac{(\sqrt n)^3}{6}
\gt\frac{(\sqrt n)^3}{6}
=\frac{n^{3/2}}{6}
$
so that
$\sum^N_{n=1} \frac 1 {e^{\sqrt n}}
\lt \sum^N_{n=1} \frac 6 {n^{3/2}}
$
which converges by the comparison test.
Note that this method
can show that
$\sum^N_{n=1} \frac 1 {e^{n^a}}
$
converges
for any $a > 0$
by going far enough out
in the power series:
$e^{n^a}
\gt \frac{(n^a)^m}{m!}
= \frac{n^{am}}{m!}
$,
so if we choose
$m \gt \frac1{a}$,
this shows that
the series converges.
