Check the convergence using the integral test for $\sum \limits_{n=1}^{\infty} \sqrt{n}\,2^{-\sqrt{n}}$

Question

$ 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.

Answer 1

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.

Answer 2

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 $\sum \frac 1 {n^2}$.

Answer 3

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 $$\sum_{n=1}^{\infty} \sqrt{n}\,2^{-\sqrt{n}}\leq C\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}<+\infty.$$

Answer 4

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 $C=\sum_{n=1}^{3}\sqrt{n} 2^{-\sqrt{n}}\leq 2$.