# Estimate $\frac 1 {a+b} + \frac 1 {(a+b)(a+2b)} + \cdots + \frac 1 {(a+b)\cdots(a+kb)} + \cdots$ [closed]

Suppose $$a>1,b>0$$ are real numbers. Consider the summation of the infinite series: $$S=\frac 1 {a+b} + \frac 1 {(a+b)(a+2b)} + \cdots + \frac 1 {(a+b)\cdots(a+kb)} + \cdots$$ How can I give a tight estimation on the summation? Apparently, one can get the upper bound: $$S\le \sum_{k=1} ^\infty \frac 1 {(a+b)^k}=\frac 1 {a+b-1}$$ But it is not tight enough. For example, fix $$a\rightarrow 1$$, and $$b=0.001$$, then $$S=38.969939$$, it seems that $$S=O(\sqrt{1/b})$$. Another example: $$a=1$$, and $$b=0.00001$$,$$S=395.039235$$.

## closed as off-topic by Jochen Wengenroth, user44191, Davide Giraudo, Jan-Christoph Schlage-Puchta, Alexey UstinovMar 18 at 12:41

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• This is not a question at the research level. Ask it at math.stackexchange.com . – user64494 Mar 14 at 7:35
• How about (1/(a+b))(a+2b)/(a+2b-1)? Gerhard "That Gets It Even Closer" Paseman, 2019.03.14. – Gerhard Paseman Mar 14 at 7:37
• I have added an example. This is still not tight because it is linear in $b$. – zbh2047 Mar 14 at 7:50
• An obvious simple upper bound is given by setting $a=0$, that is $e^{1/b}-1$. – Yaakov Baruch Mar 14 at 12:27
• If $b$ is small compared to $a$, you can write $(a+b)(a+2b)\cdots(a+kb)$ as $\exp(-\sum\log(a+ib))$, and use $\log(1+x)<x-\frac{x^2}{2}$. This yields $\mathcal{O}(b^{-1/2})$, but getting the right constants will be a bit of work. – Jan-Christoph Schlage-Puchta Mar 15 at 18:45

Maple produces a closed-form expression for the sum under consideration by

sum(1/product(b*j+a, j = 1 .. k), k = 0 .. infinity))/(a+b)  assuming a>1,b>0;


$${\frac {a{{\rm e}^{{b}^{-1}}}}{a+b}{b}^{{\frac {a-b}{b}}} \left( - \Gamma \left( {\frac {a}{b}},{b}^{-1} \right) +\Gamma \left( {\frac {a }{b}} \right) \right) }$$

Sum[1/Product[b*j + a, {j, 1, k}],{k, 0, Infinity},Assumptions -> b > 0 && a > 1]/(a + b)

• Asymptotically, as $a/b \to \infty$, this is bounded by $(a/e)^{a/b}$. – Gerald Edgar Mar 14 at 13:10
• Mathematica 11.3 confirms $S=O(\sqrt{1/b}).$ – user64494 Mar 16 at 15:45