# Upper bound on a series: $\sum_{k = M}^{\infty} (a/\sqrt{k})^k$

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

Using

$$k!\le k^k\iff\frac1{k^k}\le\frac1{k!},$$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.

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

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

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

• 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. Feb 9, 2018 at 0:27
• @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 Feb 9, 2018 at 0:31
• 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. Feb 9, 2018 at 0:33
• @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... Feb 9, 2018 at 0:37
• 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. Feb 9, 2018 at 0:42