Several ways to prove that $\sum\limits^\infty_{n=1}\left(1-\frac1{\sqrt{n}}\right)^n$ converges 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&=\left(1-\frac{1}{\sqrt{n}}\right)^n \\&=\exp\ln\left(1-\frac{1}{\sqrt{n}}\right)^n\\&=\exp \left[n\ln\left(1-\frac{1}{\sqrt{n}}\right)\right] \\&=\exp\left[ -n\sum^{\infty}_{k=1}\frac{1}{k}\left(\frac{1}{\sqrt{n}}\right)^k\right]\\&=\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]\\&=\exp \left[-\sqrt{n}-\frac{1}{2}-\sum^{\infty}_{k=3}\frac{n}{k}\left(\frac{1}{\sqrt{n}}\right)^k\right]\\&\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)<\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}<\infty,\end{align}
and so, $\sum^{\infty}_{n=1}b_n<\infty\implies \sum^{\infty}_{n=1}a_n<\infty$ by Limit comparison test. 

Noting that $\ln(\tfrac{1}{1-x}) = x + x^2/2 + O(x^3)$,
\begin{align*}
\frac{\left(\,1-\frac{1}{\sqrt{n}}\,\right)^n}{e^{-\sqrt{n}}}
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&= \frac{\exp \left(\, -n\ln \frac{1}{1-\frac{1}{\sqrt{n}}}\,\right)}{e^{-\sqrt{n}}}\\
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&= \frac{\exp \left(\, -n\left(\tfrac{1}{\sqrt{n}} + \tfrac{1}{2n} + O\left(\tfrac{1}{n^{3/2}}\right) \right)\,\right)}{e^{-\sqrt{n}}}\\
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&= e^{-1/ 2} \cdot \underbrace{e^{-O\left(\tfrac{1}{n^{1/2}}\right)}}_{\to\,1}
\to e^{-1/ 2}
\end{align*}
So $\sum_{n=1}^\infty \left(\,1-\frac{1}{\sqrt{n}}\,\right)^n$ and $\sum_{n=1}^{\infty} e^{-\sqrt{n}}$ converge or diverge together by the limit comparison test. Given that
$$\int_1^\infty e^{-\sqrt{t}}\;dt
= \left[\, -2e^{-\sqrt{x}}(\sqrt{x}+1)\,\right]_1^\infty
= \frac{4}{e} < \infty$$
we conclude that both of the latter series then converge by the integral test.
