SOLVED: Using Simply Beautiful Art's method, I managed to find the following upper bounds (and I numerically checked them). For all $a > 0$ and $m \in \mathbb{N}$, we have: $$\sum_{k=2m+1}^{\infty} \left(\frac{a}{\sqrt{k}}\right)^k \leq \left(1+\frac{a}{\sqrt{2}}\right)\frac{a}{\sqrt{2}}e\left[e^{\frac{a^2}{2e}} - \sum_{k=0}^{m-1} \frac{\left(\frac{a^2}{2e}\right)^k}{k!}\right] \leq \left(1+\frac{a}{\sqrt{2}}\right)\frac{a}{\sqrt{2}}e^{\frac{a^2}{2e}+1}$$ The first one is very close, the second one only gets close when $a$ is large.

EDIT: Rephrased the post considerably. The original statement can be found below.

Let $M \in \mathbb{N}$, $a > 0$. I have the following (convergent) series: $$\sum_{k=M}^{\infty} \left(\frac{a}{\sqrt{k}}\right)^k$$ I would like to find a closed form upper bound on this series that shows the qualitive dependece on $a$. For example, one thing we could do is the following, as pointed out by DonAntonio (assuming $a^2 > M$):

\begin{align*} \sum_{k=M}^{\infty} \left(\frac{a}{\sqrt{k}}\right)^k &= \sum_{k=M}^{\lfloor a^2+1 \rfloor} \left(\frac{a}{\sqrt{k}}\right)^k + \sum_{k=\lfloor a^2+2 \rfloor}^{\infty} \left(\frac{a}{\sqrt{k}}\right)^k \\ &\leq \left(\frac{a}{\sqrt{M}}\right)^{\lfloor a^2+1 \rfloor} \cdot (a^2+2-M) + \left(\frac{a}{\sqrt{\lfloor a^2 + 2 \rfloor}}\right)^{\lfloor a^2 + 2\rfloor} \cdot \frac{1}{1 - \frac{a}{\sqrt{\lfloor a^2 + 2 \rfloor}}} \end{align*} But this bound is not very strict, and hence does not tell us much about how the value of the series depends on $a$. So, I am looking for tighter upper bounds that reveal more of the qualitative dependence of the value of the series on $a$.

Let $M \in \mathbb{N}$, $a > 0$. I'm looking for an upper bound on the following series (it is not hard to see that it converges): $$\sum_{k=M}^{\infty} \left(\frac{a}{\sqrt{k}}\right)^k$$ However, I'm horribly stuck. Any method to upper bound this without losing too much precision would be greatly appreciated. In particular, I am interested how the resulting value depends on $a$, so an upper bound in big-O-notation in $a$ would be perfect.

I already tried to cut this series into several consecutive geometric series: \begin{align*} \sum_{k=M}^\infty \left(\frac{a}{\sqrt{k}}\right)^k &= \sum_{\ell=0}^{\infty} \sum_{k=2^\ell M}^{2^{\ell+1}M - 1} \left(\frac{a}{\sqrt{2^{\ell+1}M}}\right)^k = \sum_{\ell=0}^{\infty} \left(\frac{a}{\sqrt{2^{\ell+1}M}}\right)^{2^{\ell}M} \cdot \frac{1 - \left(\frac{a}{\sqrt{2^{\ell+1}M}}\right)^{2^\ell M}}{1 - \left(\frac{a}{\sqrt{2^{\ell+1}M}}\right)} \end{align*} But I don't see any way to continue from here.


2 Answers 2



$$k!\le k^k\iff\frac1{k^k}\le\frac1{k!},$$Assuming $M=2n$ is even and $M^2>a$, we have

\begin{align}\sum_{k=M}^\infty\left(\frac a{\sqrt k}\right)^k&=\sum_{k=M}^\infty\frac{(a\sqrt2)^k}{(k/2)^{k/2}}\\&=2\sum_{k=n}^\infty\frac{(2a^2)^k}{k^k}\\&\le2\sum_{k=n}^\infty\frac{(2a^2)^k}{k!}\\&=2e^{2a^2}-2e_{n-1}(2a^2)\\&=2e^{2a^2}\left(1-\frac{\Gamma(n,2a^2)}{\Gamma(n)}\right)\\&=2e^{2a^2}\frac{\gamma(n,2a^2)}{(n-1)!}\end{align}

using the exponential sum function and incomplete gamma functions.

  • $\begingroup$ Nice! One can do even better with this right? $$k! \leq k^ke^{-k}\sqrt{2\pi k} \Rightarrow \frac{1}{k^k} \leq \frac{e^k}{k!\sqrt{2\pi k}} \leq \frac{e^k}{k!\sqrt{2\pi M}}$$ $\endgroup$
    – arriopolis
    Feb 9, 2018 at 1:10
  • $\begingroup$ Sure, though I'm too lazy/you get the idea/affect of increasing $a$. You can get a lower bound using $k!\ge(k/e)^k$ $\endgroup$ Feb 9, 2018 at 1:11

Whenever $\;k>a^2\;$ ,we'll get that

$$\sqrt k>a\implies\text{ there exists }\;\;0<c<1\;\;s.t.\;\; \frac a{\sqrt k}<c\implies\left(\frac a{\sqrt k}\right)^k<c^k$$

  • $\begingroup$ Yes, this proves that the series converges. But if $a$ gets very large, then $k$ will have to be even larger for this to hold, so for large $a$ the biggest contribution to the result is not bounded by this relation. $\endgroup$
    – arriopolis
    Feb 9, 2018 at 0:27
  • $\begingroup$ @arriopolis $\;a\;$ can get as large as it wants, but it is a constant, so the sum's general term still is bounded as $\;k\;$ does tend to infinity $\endgroup$
    – DonAntonio
    Feb 9, 2018 at 0:31
  • $\begingroup$ Yes, but I am interested in how the sum of the series depends on $a$. I agree that it converges, but that is not what I'm after. $\endgroup$
    – arriopolis
    Feb 9, 2018 at 0:33
  • 1
    $\begingroup$ @arriopolis Then I simply don't succeed to understand you. You asked specifically for an upper bound...and now you have it . You just have to add minimal details, like $$\sum_{k=M}^\infty\left(\frac a{\sqrt k}\right)^k=\sum_{k=M}^{\lfloor a^2+1\rfloor}\left(\frac a{\sqrt k}\right)^k+\sum_{k=\lfloor a^2+2\rfloor}^\infty\left(\frac a{\sqrt k}\right)^k$$ Then the first sum in the right hand above is finite (and thus bounded...), whereas the second one is bounded by the convergent geometric series in my answer... $\endgroup$
    – DonAntonio
    Feb 9, 2018 at 0:37
  • $\begingroup$ Hmm, I think what I meant to say is that I am interested in an upper bound that has an easy qualitative dependence on $a$. For example, what happens to the upper bound when I make $a$ twice as large? How does that influence the result? This question is difficult to answer with this form of the upper bound. I will rephrase the post to indicate this more clearly. $\endgroup$
    – arriopolis
    Feb 9, 2018 at 0:42

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