Does $\sum q^\sqrt n$ converge? 
Does $\sum q^\sqrt n$ converge? ($q>0$)

It is clear that if $q\ge1$ series diverges, but what about $q\in(0; 1)$?
 A: For $q\in(0,1)$:
$$\sum_{n=1}^\infty q^{\sqrt n}<\sum_{n=1}^\infty q^{\lfloor\sqrt n\rfloor}\stackrel*=\sum_{k=1}^\infty (2k+1)q^k<\infty$$
$(*)$ Note that for each $k\in\Bbb N$ there are exactly $2k+1$ natural numbers $n$ such that $\lfloor\sqrt n\rfloor=k$.
EDIT (motivated by Clement's comment):
Let $r\in(0,1)$. Take some natural $s>1/r$. For each natural $k$, there are
$$p(k)=(k+1)^s-k^s$$
natural numbers $n$ such that $\lfloor\sqrt[s]n\rfloor=k$. Then
$$\sum_{n=1}^\infty q^{n^r}<\sum_{k=1}^\infty p(k)q^k<\infty$$
A: Yes, it does converge (and quite fast, but that's besides the point). We can prove it by comparison with, for instance, the series $\sum_{n=1}^\infty \frac{1}{n^2}$.
Let $q\in(0,1)$ be any number.
$$\frac{q^{\sqrt{n}}}{\frac{1}{n^2}} = n^2 e^{-\sqrt{n} \ln\frac{1}{q}}
= e^{-\sqrt{n} \ln\frac{1}{q}+2\ln n}
= e^{-\ln\frac{1}{q}\left( \sqrt{n} - \frac{2}{\ln\frac{1}{q}}\ln n\right)}$$
Now, you can use the fact$^{(\dagger)}$ that for any real number $a\in\mathbb{R}$,
$$
\sqrt{n}-a\ln n \xrightarrow[n\to\infty]{} \infty
$$
to conclude that $\frac{q^{\sqrt{n}}}{\frac{1}{n^2}}\xrightarrow[n\to\infty]{} 0$.

If $(\dagger)$ is not obvious or known to you:
$$
\sqrt{n}-a\ln n = \sqrt{n}\left(1-a\frac{\ln n}{\sqrt{n}}\right) 
= \sqrt{n}\left(1-2a\frac{\ln \sqrt{n}}{\sqrt{n}}\right) $$
so it is sufficient to know or prove that $\frac{\ln x}{x} \xrightarrow[x\to\infty]{} 0$.
A: Let $f:\mathbb{Z}_{\geq 0}\to\mathbb{R}_{\geq 0}$ be a function such that, as a function of $m\in\mathbb{Z}_{\geq 0}$, the cardinality of the preimage $T_m:=f^{-1}\big([m,m+1)\big)$ is polynomially bounded, namely, for some polynomial $p(X)\in\mathbb{R}[X]$, $\big|T_m\big|\leq p(m)$ for every $m$.  (In fact, we only need that $T_m$ is finite for all $m$ and $\limsup\limits_{m\to\infty}\,\sqrt[m]{\left|T_m\right|}\leq1$.)  Then, for $q>0$, the infinite sum $\displaystyle\sum_{n=0}^\infty\,q^{f(n)}$ converges if and only if $q<1$.  
It is trivial that $q<1$ is necessary for the sum to converge.  We shall show that the sum indeed converges when $q<1$.  Observe that
$$\sum_{n=0}^\infty\,q^{f(n)}=\sum_{m=0}^\infty\,q^{m}\,\sum_{j\in T_m}\,q^{f(j)-m}\leq\sum_{m=0}^\infty\,q^m\,\left|T_m\right|\,.$$
Since $\left|T_m\right|$ is polynomially bounded, we conclude that $\sum\limits_{m=0}^\infty\,q^m\,\left|T_m\right|<\infty$ for $q\in(0,1)$.
In particular, if $f(n)=\sqrt{n}$, then $\big|T_m\big|=(m+1)^2-m^2=2m+1$ is linear, whence polynomially bounded.  Hence, for $q>0$, the sum $\sum\limits_{n=0}^\infty\,q^{\sqrt{n}}$ is convergent iff $q<1$.
A: *

*Write $q=\frac{1}{1+y}$

*Note that $(1+y)^k > {k \choose 3}y^3$

*Deduce that $q^k < \frac{y^{-3}3!}{k(k-1)(k-2)}$

*Take $k=\sqrt{n}$ to see$$q^{\sqrt{n}}=O(n^{-3/2})$$

*Remark that $n^{-3/2 }$ has a convergent sum

*Deduce that the sum is convergent

A: We can use the estimate $\sqrt n\geq\ln n$ to get
$$
\sum_{n=1}^\infty q^{\sqrt n}\leq\sum_{n=1}^\infty q^{\ln n}=\sum_{n=1}^\infty n^{\ln q}<\infty,\ \ \text{ if }\ln q<-1,$$
which happens for $0<q<\frac{1}{e}\approx0.367878$. We have thus obtained only a partial solution, but this argument can be modified  in order to get the desired result:
Let's fix $k\geq1$. As $\frac{\sqrt n}{k\ln n}\to\infty$ when $n\to\infty$ we know that there is some $N(k)\in\mathbb{N}$ such that $$\sqrt{n}\geq k\ln n,\ \ \text{for }n\geq N.$$ Performing the above estimations again we arrive at
$$
\sum_{n=N}^\infty q^{\sqrt n}\leq\sum_{N}^\infty n^{k\ln q}<\infty,
\ \ \text{ if }k\ln q<-1,$$ which happens for $0<q<e^{-\frac{1}{k}}$. As $e^{-\frac{1}{k}}\nearrow1$ when $k\to\infty$ this yields the desired convergence for $0<q<1$. 
