I am trying to investigate the convergence of the series: \begin{equation} \sum_{n=1}^{\infty} \bigg ( 1 -\frac{1}{\sqrt{n}} \bigg )^n. \end{equation} According to the Mathematica, the series converges, but I am unable to come up with a series with which I can compare the series above etc. Any leads would be appreciated.
I believe there are several ways to prove that $\sum\limits^{\infty}_{n=1}\left(1-\frac{1}{\sqrt{n}}\right)^n$ converges. Can you, please, post yours so that we can learn from you?  HERE IS ONE   Let $n\in\Bbb{N}$ be fixed such that $a_n=\left(1-\frac{1}{\sqrt{n}}\right)^n.$ Then, \begin{align} a_n&amp;=\left(1-\frac{1}{\sqrt{n}}\right)^n \\&amp;=\exp\ln\left(1-\frac{1}{\sqrt{n}}\right)^n\\&amp;=\exp \left[n\ln\left(1-\frac{1}{\sqrt{n}}\right)\right] \\&amp;=\exp\left[ -n\sum^{\infty}_{k=1}\frac{1}{k}\left(\frac{1}{\sqrt{n}}\right)^k\right]\\&amp;=\exp\left[ -n\left(\frac{1}{\sqrt{n}}+\frac{1}{2n}+\sum^{\infty}_{k=3}\frac{1}{k}\left(\frac{1}{\sqrt{n}}\right)^k\right)\right]\\&amp;=\exp \left[-\sqrt{n}-\frac{1}{2}-\sum^{\infty}_{k=3}\frac{n}{k}\left(\frac{1}{\sqrt{n}}\right)^k\right]\\&amp;\equiv\exp \left(-\sqrt{n}\right)\exp \left(-\frac{1}{2}\right)\end{align} Choose $b_n=\exp \left(-\sqrt{n}\right)$, so that  \begin{align} \dfrac{a_n}{b_n}\to\exp \left(-\frac{1}{2}\right).\end{align} Since $b_n \to 0$, there exists $N$ such that for all $n\geq N,$ \begin{align} \exp \left(-\sqrt{n}\right)&lt;\dfrac{1}{n^2}.\end{align} Hence, \begin{align}\sum^{\infty}_{n=N}b_n= \sum^{\infty}_{n=N}\exp \left(-\sqrt{n}\right)\leq \sum^{\infty}_{n=N}\dfrac{1}{n^2}&lt;\infty,\end{align} and so, $\sum^{\infty}_{n=1}b_n&lt;\infty\implies \sum^{\infty}_{n=1}a_n&lt;\infty$ by Limit comparison test.
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