Any formula for $\sum_{k=0}^{n} e^{(kt)^m}$? I wonder if there is a formula for $\sum_{k=0}^{n} e^{k^mt^m}$ or non-trivial estimate, where we may assume $t=o(1)$ and $m\ge 2$ is a positive integer?
 A: It's easiest to break this into cases depending on whether $t$ is positive, zero, or negative and, if negative, whether $|t| \in (0,1)$ or $|t| > 1$, and also whether $m$ is even or odd.


*

*$t < 0$, $|t| \geq 1$, $m$ odd : The summands decrease superexponentially, dominated by the first term.  The sum is in $[1,3)$.  (We could tighten the upper bound to $2 + \mathrm{e}^{-7}$ easily, but this seems excessive.)

*$t < 0$, $|t| < 1$, $m$ odd : Lower bounded by the first term, $1$, upper bounded by $n+1$.

*$t < 0$ and $m$ is even.  Same as $t > 0$, below.

*$t = 0$ : The sum is $n+1$.

*$t > 0$ : The terms grow superexponentially.  The sum is dominated by its last term and is $C \mathrm{e}^{(n t)^m}$ for some $C \in [1,2]$.

Regarding the question in comments about the last case.
\begin{align*}
\sum_{k=0}^n \mathrm{e}^{(kt)^m}
    &= \left( \sum_{k=0}^{n} \mathrm{e}^{(kt)^m-(nt)^m}\right) \mathrm{e}^{(nt)^m}  \\
    &= \left( \sum_{k=0}^{n} \mathrm{e}^{-(n^m - k^m)t^m}\right) \mathrm{e}^{(nt)^m}
\end{align*}
I should have believed me when I said the relative magnitudes of $t$ and $1$ matter.


*

*$t > 0$, $|t| < 1$: The $k=n$ term in the parentheses above gives a lower bound of $\mathrm{e}^{(nt)^m}$.  Every term in the parenthesized sum is bounded between $0$ and $1$, so your sum is bounded by $(n+1)\mathrm{e}^{(nt)^m}$.

*$t > 0$, $|t| \geq 1$: The sum in parentheses is \begin{align*}  
&\left( 1 + \left( \mathrm{e}^{-t^m} \right)^{n^m - (n-1)^m} + \left( \mathrm{e}^{-t^m} \right)^{n^m - (n-2)^m} + \cdots + \left( \mathrm{e}^{-t^m} \right)^{n^m - 0} \right) \\ 
&\qquad < \left( 1 + \left( \mathrm{e}^{-t^m} \right) + \left( \mathrm{e}^{-t^m} \right)^2 + \cdots + \left( \mathrm{e}^{-t^m} \right)^n \right)  \\
&\qquad < \frac{1}{1 - \mathrm{e}^{-t^m}}  \\
&\qquad = 1 + \frac{\mathrm{e}^{-t^m}}{1 - \mathrm{e}^{-t^m}}  \\
&\qquad \in \left[1,\frac{\mathrm{e}}{\mathrm{e}-1} \right]  \\
&\qquad \subset [1,2]  \text{.}
\end{align*}  Your sum is $C \mathrm{e}^{(n t)^m}$ for some $C \in [1,2]$.

