Check the convergence using the integral test for $\sum \limits_{n=1}^{\infty} \sqrt{n}\,2^{-\sqrt{n}}$ $ 2^{-\sqrt{x}}$ is positive, decreasing, continuous, so we can use the integral test.
$\int \limits_{1}^{\infty} \sqrt{x} \, 2^{-\sqrt{x}} = \lim \limits_{T \to \infty}  \int \limits_{1}^{T} \sqrt{x} \, 2^{-\sqrt{x}} dx$.
And now, how to calculate $\int \limits_{1}^{T} \sqrt{x} \, 2^{-\sqrt{x}} dx$?
Substitution $u = \sqrt{x}$ obviously does not work.
 A: With the substitution $u = \sqrt{x}$, we find
$$
\int_1^T \sqrt{x}\,2^{-\sqrt{x}}dx = 
\int_{1}^\sqrt{T} u\,2^{-u} \ 2\sqrt{x}\ du = 
\int_{1}^\sqrt{T} u\,2^{-u} \ 2u\ du = 
2\int_{1}^\sqrt{T} u^2 2^{-u}
$$
This integral can be done using integration by parts.  We find that the improper integral converges.
A: Actually, it is not necessary to evaluate that integral exactly. Note that for $x> 0$, then
$$2^x=\exp(x\ln(2))\geq \frac{(x\ln(2))^4}{4!}\implies x2^{-x}\leq \frac{C}{x^3} $$
where $C=24/\ln^4(2)$.
Hence
$$\sum_{n=1}^{\infty} \sqrt{n}\,2^{-\sqrt{n}}\leq C\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}<+\infty.$$
A: As an alternative note that
$$\frac{\frac{\sqrt{n}} {2^{\sqrt{n}}}}{\frac 1 {n^2}}\to 0$$
then $\sum \frac{\sqrt{n}} {2^{\sqrt{n}}}$ converges by limit comparison test with $\sum \frac 1 {n^2}$.
A: Actually you do not even need integrals. Since $\sqrt{n}\,2^{-\sqrt{n}}$ is decreasing for $n\geq 2$,
$$ \sum_{n=1}^{M^2-1}\sqrt{n}\,2^{-\sqrt{n}}=C+\sum_{m=2}^{M-1}\sum_{n=m^2}^{(m+1)^2-1}\sqrt{n} 2^{-\sqrt{n}}\leq C+\sum_{m=2}^{M-1}2m^2 2^{-m}\leq C+\sum_{m\geq 2}2m^2 2^{-m}=C+11 $$
where $C=\sum_{n=1}^{3}\sqrt{n} 2^{-\sqrt{n}}\leq 2$.
