Good upper bound for $\Gamma(1-b,\log(a))-\Gamma(1-b,N\log(a))$, where $a,b \in (0, 1)$ and $N \ge 1$

Question

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 for more details.

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

Answer 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}

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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}