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

It is clear that if $q\ge1$ series diverges, but what about $q\in(0; 1)$?

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  • $\begingroup$ $q^{\sqrt{n}}=\exp(\ln(q) \sqrt{n})$. Since $\ln(q)<0$, $\exp(\ln(q) \sqrt{n})=o\left( \frac{1}{n^2}\right)$ for example, which yields convergence. $\endgroup$ Commented Jan 22, 2017 at 20:27

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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 $$\sum_{n=1}^\infty q^{n^r}<\sum_{k=1}^\infty p(k)q^k<\infty$$

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    $\begingroup$ (+1) Slick proof. (It does not easily generalize to arbitrary $r\in(0,1)$ instead of $1/2$ for $n^r$, but as one-liners go it is quite great.) $\endgroup$
    – Clement C.
    Commented Jan 22, 2017 at 20:37
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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 $\frac{\ln x}{x} \xrightarrow[x\to\infty]{} 0$.

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  • 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
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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 $q<1$.

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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 $0<q<1$.

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