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: $$S\le \sum_{k=1} ^\infty \frac 1 {(a+b)^k}=\frac 1 {a+b1}$$ 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 offtopic by Jochen Wengenroth, user44191, Davide Giraudo, JanChristoph SchlagePuchta, Alexey Ustinov Mar 18 at 12:41
This question appears to be offtopic. The users who voted to close gave this specific reason:
 "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Jochen Wengenroth, user44191, Davide Giraudo, Alexey Ustinov

$\begingroup$ This is not a question at the research level. Ask it at math.stackexchange.com . $\endgroup$ – user64494 Mar 14 at 7:35

$\begingroup$ How about (1/(a+b))(a+2b)/(a+2b1)? Gerhard "That Gets It Even Closer" Paseman, 2019.03.14. $\endgroup$ – Gerhard Paseman Mar 14 at 7:37

$\begingroup$ I have added an example. This is still not tight because it is linear in $b$. $\endgroup$ – zbh2047 Mar 14 at 7:50

$\begingroup$ An obvious simple upper bound is given by setting $a=0$, that is $e^{1/b}1$. $\endgroup$ – Yaakov Baruch Mar 14 at 12:27

$\begingroup$ 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. $\endgroup$ – JanChristoph SchlagePuchta Mar 15 at 18:45
Maple produces a closedform 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;
$${\frac {a{{\rm e}^{{b}^{1}}}}{a+b}{b}^{{\frac {ab}{b}}} \left(  \Gamma \left( {\frac {a}{b}},{b}^{1} \right) +\Gamma \left( {\frac {a }{b}} \right) \right) } $$
Addition. So does Mathematica through
Sum[1/Product[b*j + a, {j, 1, k}],{k, 0, Infinity},Assumptions > b > 0 && a > 1]/(a + b)

$\begingroup$ Asymptotically, as $a/b \to \infty$, this is bounded by $(a/e)^{a/b}$. $\endgroup$ – Gerald Edgar Mar 14 at 13:10


$\begingroup$ @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. 2 days ago