Let $a,b \in (0, 1)$ and $N \ge 1$, and consider the incomplete gamma function $x \mapsto \Gamma(1-a,x)$.


Question
========


Is there a simple bound (involving 'simple function's) for the expression $\Gamma(1-b,\log(a))-\Gamma(1-b,N\log(a))$ ?


Motivation
==========


Ultimately, I'm interesting in bounding the sum $S\_N:=\sum\_{n=1}^N a^{-n} n^{-b}$, via the sum-integral inequality, I though of bounding the corresponding integral instead. See this [SE question](https://math.stackexchange.com/questions/3198696/good-upper-bound-for-sum-n-1n-a-n-n-b-for-a-b-in-0-1?noredirect=1#comment6583028_3198696) for more details.


According to wolfram alpha,
$$\int\_{1}^{N} a^{-t}t^{-b}dt = \log^{b-1}(a)\left(\Gamma(1-b,\log(a))-\Gamma(1-b,N\log(a))\right),
$$



Good upper bound for $\Gamma(1-b,\log(a))-\Gamma(1-b,N\log(a))$, where $a,b \in (0, 1)$ and $N \ge 1$
Let us go for your ultimate goal and provide a tight upper bound on 
\begin{equation}
 s:=\sum\_{n=1}^N a^{-n} n^{-b}=\sum\_{n=1}^N c^n n^{-b},
\end{equation}
where $c:=1/a>1$ and $b>0$. We assume that $N\to\infty$. Take any natural $M$ such that $1<M<N$ and write 
\begin{equation}
 s=s\_1+s\_2,
\end{equation}
where 
\begin{equation}
 s\_1:=\sum\_{n=1}^M c^n n^{-b}\le Mc^M
\end{equation}
and 
\begin{align\*}
 s\_2&:=\sum\_{n=M+1}^N c^n n^{-b} \\ 
& =\sum\_{n=M+1}^N c^N N^{-b} \prod\_{j=n}^{N-1}\Big(\frac1c\Big(\frac{j+1}{j}\Big)^b\Big) \\ 
 &\le\sum\_{n=M+1}^N c^N N^{-b} \Big(\frac1c\Big(\frac{M+1}{M}\Big)^b\Big)^{N-n} \\ 
 &\le\sum\_{n=-\infty}^N c^N N^{-b} \Big(\frac1c\Big(\frac{M+1}{M}\Big)^b\Big)^{N-n} \\ 
 &=\frac{c^N N^{-b}}{1-\frac1c\Big(\frac{M+1}{M}\Big)^b} \\ 
 &\le\frac{c^N N^{-b}}{1-\frac{1+b/M}c}
\end{align\*}
So, 
\begin{equation}
 s\le B:= Mc^M+\frac{c^N N^{-b}}{1-\frac{1+b/M}c}. 
\end{equation}
Choosing now $M$ so that 
\begin{equation}
 N-M\sim t\log\_c N
\end{equation}
for any fixed real $t>b+1$, we see that 
\begin{equation}
 B\sim\frac{c^N N^{-b}}{1-\frac1c}. 
\end{equation}




---


The upper bound $B$ on $s$ is tight, because  

\begin{equation}
 s\ge\sum\_{n=1}^N c^n N^{-b}
 \sim\frac{c^N N^{-b}}{1-\frac1c}\sim B. 
\end{equation}


