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


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


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

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

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}

  • $\begingroup$ Interesting, thanks. BTW, can't the whole thing be non asymptotic by just taking $N - 2(b+1)\log_c N < M < N- (b+1)\log_c N$ ? $\endgroup$ – dohmatob Apr 25 '19 at 13:02
  • $\begingroup$ @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. $\endgroup$ – Iosif Pinelis Apr 25 '19 at 19:38
  • $\begingroup$ Ok, thanks again. I like your approach "OK, why do we not just solve the harder problem right away [...] Done." :) $\endgroup$ – dohmatob Apr 26 '19 at 12:59

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