Does $\sum_{n=1}^{\infty} \frac1{2^{\sqrt{n}}}$ converge $$\sum_{n=1}^{\infty} \frac1{2^{\sqrt{n}}}$$
Here are my ideas $$\frac1{2^{\sqrt{n}}}\gt \frac1{2^{n}}$$
and
$$\frac1{2^{n}}\gt\frac1{n!}$$
But i don't know what to do with them. Anyone has any ideas?
 A: The integral test tell us that $$\int_{1}^{\infty}\frac{1}{2^{\sqrt{t}}}dt\stackrel{u=\sqrt{t}}{=}2\int_{1}^{\infty}\frac{u}{2^{u}}du
 $$ $$\stackrel{IBP}{=}\frac{1}{\log\left(2\right)}+\frac{2}{\log\left(2\right)}\int_{1}^{\infty}2^{-u}du=\frac{1+\log\left(2\right)}{\log^{2}\left(2\right)}$$ so the series converges.
A: It's
$\displaystyle 0<\sum_{n=1}^{\infty} \frac{1}{2^{\sqrt{n}}}<\sum_{n=1}^{\infty} \frac{2n+1}{2^n}=2\cdot 2 +1=5$
because of $\displaystyle \sum\limits_{n=m^2}^{(m+1)^2-1}\frac{1}{2^{\sqrt{n}}}<\frac{2m+1}{2^\sqrt{m^2}}$ for $m\in\mathbb{N}$.
A: This is essentially what @user90369 wrote, shown a bit differently:
$$\sum_{n=1}^{\infty} \frac1{2^{\sqrt{n}}}=\\\underbrace{\frac1{2^{1}}+\frac1{2^{\sqrt2}}+\frac1{2^{\sqrt3}}}_{3\rm{ \, terms}}+\underbrace{\frac1{2^{2}}+\cdots+\frac1{2^{\sqrt8}}}_{5\rm{ \, terms}}+\underbrace{\frac1{2^{3}}+\cdots+\frac1{2^{\sqrt{15}}}}_{7\rm{ \, terms}}+\cdots\\< \frac3{2^1}+\frac5{2^2}+\frac7{2^3}+\cdots.$$
A: Hint: Try using Cauchy's root test. 
Let $a_{n}=\frac{1}{2^\sqrt{n}}$, then $a_{n}^\frac{1}{n}=\dfrac{1}{2^\frac{1}{\sqrt{n}}}=2^{-\frac{1}{\sqrt{n}}}$. Now, convince yourself that $a_{n}^\frac{1}{n}$ is lesser than $2^{-\frac{1}{n}}$, which in turn is less than $1$. This means, by Cauchy's root test that $\exists$ k $\in$ $\mathbb R$ such that $a_{n}^\frac{1}{n}<k<1$, and hence the series converges.
A: Claim: $1/2^{\sqrt n} \le 1/n^2$ for large $n.$ If the claim is true, then our series converges by the comparison test. Proof of claim: It's the same as saying $\sqrt n\ln (1/2) \le -2\ln n,$ or $\sqrt n \ge (-2/\ln (1/2))\ln n.$ The last is true for large $n,$ since any positive power of $n$ eventually overtakes any positive constant times $\ln n.$
