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 $b=0.00001$,$S=395.039235$.
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$\begingroup$ This is not a question at the research level. Ask it at math.stackexchange.com . $\endgroup$– user64494Commented Mar 14, 2019 at 7:35
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$\begingroup$ How about (1/(a+b))(a+2b)/(a+2b-1)? Gerhard "That Gets It Even Closer" Paseman, 2019.03.14. $\endgroup$– Gerhard PasemanCommented Mar 14, 2019 at 7:37
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$\begingroup$ I have added an example. This is still not tight because it is linear in $b$. $\endgroup$– zbh2047Commented Mar 14, 2019 at 7:50
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$\begingroup$ An obvious simple upper bound is given by setting $a=0$, that is $e^{1/b}-1$. $\endgroup$– Yaakov BaruchCommented Mar 14, 2019 at 12:27
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$\begingroup$ 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. $\endgroup$– Jan-Christoph Schlage-PuchtaCommented Mar 15, 2019 at 18:45
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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;
$${\frac {a{{\rm e}^{{b}^{-1}}}}{a+b}{b}^{{\frac {a-b}{b}}} \left( - \Gamma \left( {\frac {a}{b}},{b}^{-1} \right) +\Gamma \left( {\frac {a }{b}} \right) \right) } $$
Addition. So does Mathematica through
Sum[1/Product[b*j + a, {j, 1, k}],{k, 0, Infinity},Assumptions -> b > 0 && a > 1]/(a + b)
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$\begingroup$ Asymptotically, as $a/b \to \infty$, this is bounded by $(a/e)^{a/b}$. $\endgroup$ Commented Mar 14, 2019 at 13:10
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$\begingroup$ Mathematica 11.3 confirms $ S=O(\sqrt{1/b}).$ $\endgroup$ Commented Mar 16, 2019 at 15:45
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$\begingroup$ @user64494: You have recently tried to have a minor edit validated, but it got rejected. Immediately after, you submitted the same edit attempt once more. Please don't do this! It is not polite, it feels aggressive and it leaves the impression that you are desperately begging for those +2 reputation points that you get for each approved edit. $\endgroup$– Alex M.Commented Mar 20, 2019 at 18:07