# 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 $$a,b \in (0, 1)$$ 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, $$\int_{1}^{N} a^{-t}t^{-b}dt = \log^{b-1}(a)\left(\Gamma(1-b,\log(a))-\Gamma(1-b,N\log(a))\right),$$

• If $|(n-1)\log(a)|<1$, you can get an elementary upper bound for the integral from $$\int_1^N \frac{1-(t-1)\log(a)+(t-1)^2\log^2(a)/2}{a}t^{-b}dt$$ – Matt F. Apr 24 '19 at 19:33

Let us go for your ultimate goal and provide a tight upper bound on $$$$s:=\sum_{n=1}^N a^{-n} n^{-b}=\sum_{n=1}^N c^n n^{-b},$$$$ where $$c:=1/a>1$$ and $$b>0$$. We assume that $$N\to\infty$$. Take any natural $$M$$ such that $$1 and write $$$$s=s_1+s_2,$$$$ where $$$$s_1:=\sum_{n=1}^M c^n n^{-b}\le Mc^M$$$$ and \begin{align*} s_2&:=\sum_{n=M+1}^N c^n n^{-b} \\ & =\sum_{n=M+1}^N c^N N^{-b} \prod_{j=n}^{N-1}\Big(\frac1c\Big(\frac{j+1}{j}\Big)^b\Big) \\ &\le\sum_{n=M+1}^N c^N N^{-b} \Big(\frac1c\Big(\frac{M+1}{M}\Big)^b\Big)^{N-n} \\ &\le\sum_{n=-\infty}^N c^N N^{-b} \Big(\frac1c\Big(\frac{M+1}{M}\Big)^b\Big)^{N-n} \\ &=\frac{c^N N^{-b}}{1-\frac1c\Big(\frac{M+1}{M}\Big)^b} \\ &\le\frac{c^N N^{-b}}{1-\frac{1+b/M}c} \end{align*} So, $$$$s\le B:= Mc^M+\frac{c^N N^{-b}}{1-\frac{1+b/M}c}.$$$$ Choosing now $$M$$ so that $$$$N-M\sim t\log_c N$$$$ for any fixed real $$t>b+1$$, we see that $$$$B\sim\frac{c^N N^{-b}}{1-\frac1c}.$$$$
The upper bound $$B$$ on $$s$$ is tight, because
$$$$s\ge\sum_{n=1}^N c^n N^{-b} \sim\frac{c^N N^{-b}}{1-\frac1c}\sim B.$$$$
• Interesting, thanks. BTW, can't the whole thing be non asymptotic by just taking $N - 2(b+1)\log_c N < M < N- (b+1)\log_c N$ ? – dohmatob Apr 25 '19 at 13:02
• @dohmatob : Yes, the upper bound $B$ on $s$ is non-asymptotic. Also, with $M$ appropriately chosen, the bound $B$ is asymptotically tight, as shown in the answer. – Iosif Pinelis Apr 25 '19 at 19:38