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

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$ .

$\bigcap I_\gamma$ if $I_0 \supset \cdots \supset\I_{\gamma} \cdots$ are unbounded

Let be $ \kappa$ a regular cardinal and $ I_0 \supset \cdots \supset I_{\gamma} \cdots$ are unbounded subsets of $ \kappa$ for $ \gamma < \lambda <\kappa$ where $ \lambda$ is limit. I want to show $ \bigcap I_\gamma$ is unbounded. Is it true? It’s easy show it’s not empty for regularity of $ \kappa$ .

On constructing free action of the cyclic group $\Bbb Z/3\Bbb Z$ on $S^n \times \cdots \times S^n$($n$ is odd)

Can we construct a free action of the cyclic group $ \Bbb Z/3\Bbb Z$ on $ S^n \times \cdots \times S^n$ ($ n$ is odd) without multiplying any coordinate by $ e^{2\pi i/3} $ ? In other words, do I Need to multiply at least one coordinate by a 3rd root of unity to get a free action of $ \Bbb Z/3\Bbb Z$ ? I have tried to construct such examples but could not find any.

Thank you so much in advance.