As the title states, I'm looking for some insight into the following summation: $$ \sum_{i=1}^{n}i^{\sqrt{i}} $$ In particular, I am interested in how I could go about finding upper and lower bounds that are as close to each other as possible.
If it is possible to put the summation in terms of a simple function that doesn't involve a summation, that would be ideal but not necessary.
Not bounds as asked, but asymptotes, which may end up useful anyways.
Using something akin to a $u$ substitution or the Cauchy condensation test, one has
$$\sum_{k=1}^nk^{\sqrt k}\approx\sum_{k=1}^{\{\sqrt n\}}2k^{2k+1}$$
where $\{x\}$ is $x$ rounded to the nearest integer. It thus remains to approximate this sum, which can be done similarly as done here by considering what happens when it is divided by it's largest term:
\begin{align}\frac1{n^{2n+1}}\sum_{k=1}^nk^{2k+1}&=\sum_{k=0}^{n-1}\frac1{n^{2k}}\left(1-\frac kn\right)^{2(n-k)+1}\\&=\sum_{k=0}^{n-1}\frac1{(en)^{2k}}\left(1-\frac kn\right)\exp\left[2k\sum_{j=1}^\infty\frac{(k/n)^j}{j(j+1)}\right]\\&=\sum_{k=0}^m\frac1{(en)^{2k}}\left(1-\frac kn\right)\exp\left[2k\sum_{j=1}^{2m-2k+1}\frac{(k/n)^j}{j(j+1)}\right]+\mathcal O(n^{-(2m+2)})\end{align}
The first few terms are then given by
$$\frac1{n^{2n+1}}\sum_{k=1}^nk^{2k+1}=1+\frac1{e^2n^2}+\left(\frac1{e^4}-\frac1{6e^2}\right)\frac1{n^4}+\left(\frac2{e^4}-\frac1{6e^2}\right)\frac1{n^5}+\mathcal O(n^{-6})$$
assuming I expanded that out right...
