In PG office PhD students are preparing their "Qualifying Exam" in Advanced Calculus, one of them discusses with me the following interesting question:
Problem. Find a sequence $\{x_n\}$ of real numbers such that $\dis \limn x_n=1$ and \[
\sum_{n=1}^\infty \frac{1}{n^{x_n}}<\infty.
\]Solution.
For every $\alpha>1$ the series $\sum 1/n^{\alpha}$ converges, thus, for every $k\in \N$ and $\alpha=1+1/k$, there will be an $N_k$ such that \[
\sum_{n\ge N_k}\frac{1}{n^{1+1/k}}<\frac{1}{2^k}.
\] We may assume $\{N_k\}$ is strictly increasing, then we can define \[
x_j = 1+\frac{1}{k}\quad \text{if } N_k\leq j < N_{k+1}.
\] And of course this choice will do! (with any choices $x_1,\dots,x_{N_1-1}$ whom we don't care)$\qed$
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