<s>[INST] 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. 
 [/INST] For $n\ge2$,
$$
\begin{align}
\left(1-\frac1{\sqrt{n}}\right)^n
&=\left(\left(1+\frac1{\sqrt{n}-1}\right)^{\sqrt{n}}\right)^{-\sqrt{n}}\tag{1a}\\
&\le\left(1+\frac{\sqrt{n}}{\sqrt{n}-1}\right)^{-\sqrt{n}}\tag{1b}\\[9pt]
&\le2^{-\sqrt{n}}\tag{1c}
\end{align}
$$
Explanation:
$\text{(1a)}$: $1-\frac1{\sqrt{n}}=\left(1+\frac1{\sqrt{n}-1}\right)^{-1}$
$\text{(1b)}$: Bernoulli's Inequality
$\text{(1c)}$: $1+\frac{\sqrt{n}}{\sqrt{n}-1}\ge2$
Since $(1)$ holds for $n=1$, it is true for $n\ge1$.
$$
\begin{align}
\sum_{n=1}^\infty\left(1-\frac1{\sqrt{n}}\right)^n
&\le\sum_{n=1}^\infty2^{-\sqrt{n}}\tag{2a}\\
&=\sum_{k=1}^\infty\sum_{n=(k-1)^2+1}^{k^2}2^{-\sqrt{n}}\tag{2b}\\
&\le\sum_{k=1}^\infty(2k-1)\,2^{-k+1}\tag{2c}\\[9pt]
&=6\tag{2d}
\end{align}
$$
Explanation:
$\text{(2a)}$: apply $(1)$
$\text{(2b)}$: group the terms
$\text{(2c)}$: $2k-1$ terms of size less than $2^{-k+1}$
$\text{(2d)}$: compute sum using
$\phantom{\text{(2d):}}$ $\sum\limits_{k=1}^\infty x^k=\frac{x}{1-x}\quad$ and $\quad\sum\limits_{k=1}^\infty kx^k=\frac{x}{(1-x)^2}$
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