I'm working on this problem: $$ \mbox{Find all}\ a\ \mbox{for which}\ \sum_{k=1}^{\infty}\left(1-k^{-a}\right)^{k}\ \mbox{converges}\ ? $$
- Clearly, if $a\ge 1$ or $a < 0$, the series will diverge based on the limit of general term as $k\to \infty$.
- I believe that the series converges if $a \in \left[0,1\right)$ but I don't know how to show this.
- Well, simple numerical tests with Mathematica shows that $\left(1 - k^{-a}\right)^k$ tends to zero faster than for instance $1/k^{b}$ for any $b > 1$ so limit comparison test could be used. But how can I prove this theoretically $?$.
Any help would be greatly appreciated. Thanks.
Using the inequality $1+x \le e^x$ for all real numbers $x$, when $a\ge0$ we may write $$ 0 \le (1-k^{-a})^k \le (e^{-k^{-a}})^k = e^{-k^{1-a}}. $$ And when $a<1$, indeed $e^{-k^{1-a}}$ tends to $0$ faster than $k^{-b}$ for any $b>1$ which implies convergence of the original series. We can verify this last claim by showing that $b\log k$ tends to infinity slower than $k^{1-a}$ (that's an easy l'Hôpital's rule calculation) and then exponentiating both sides.