Answer
The series $\sum_{n=3}^{\infty}\frac{ 1}{e^{\sqrt n} }$ converges.
Work Step by Step
We have the series:
$\sum_{n=3}^{\infty}\frac{ 1}{e^{\sqrt n} }$
We apply the integral test:
$\int_1^{\infty} e^{-\sqrt{x}}dx=\int_1^{\infty} 2ue^{-u}du$
Where $u=\sqrt{x}$
We solve by integration by parts:
$\int_1^{\infty} 2ue^{-u}du=-2e^{-u}(1+u)|_1^{\infty}=\lim_{R\rightarrow\infty} -2e^{-u}(1+u)|_1^{R}=\lim_{R\rightarrow\infty}-2e^{-R}(1+R)-[-2e^{-1}(1+1)]=\lim_{R\rightarrow\infty}\frac{-2(1+R)}{e^R}+\frac{4}{e}$
This leads to an indeterminate form, so we apply L'Hopital's rule:
$=\lim_{R\rightarrow\infty}\frac{-2}{e^R}=0$
Since the integral converges, then the series converges by the integral test.
