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 $b=0.00001$,$S=395.039235$.



Estimate $\frac 1 {a+b} + \frac 1 {(a+b)(a+2b)} + \cdots + \frac 1 {(a+b)\cdots(a+kb)} + \cdots$
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)

```

