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

Question

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.

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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 $a$.

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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.

Answer 1

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.

Answer 2

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

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