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. 
 A: Here is an elementary approach, which has the advantage of being based, first, on a basic inequality which is so useful that one should keep it in mind anyway, second, on a condensation technique which is so useful that one should keep it in mind anyway, and third, on a standard series which is so useful that one should keep it in mind anyway as well... 
A basic inequality: For every $x$,

$$1-x\leqslant e^{-x}\tag{1}$$ 

(This stems, for example, from the fact that, the exponential function being convex, its graph is above its tangent at $x=0$.)
Now, we massage slightly the Uhr inequality $(1)$ above: if both sides are nonnegative, the inequality is preserved when raised to any positive power, hence, for every $x\leqslant1$ and every nonnegative $n$, $$(1-x)^n\leqslant e^{-nx}$$ for example, for every positive $n$, $$\left(1-\frac1{\sqrt n}\right)^n\leqslant e^{-\sqrt n}$$ hence the series of interest converges as soon as the series $$\sum_ne^{-\sqrt n}$$ converges. 
A condensation technique: In words, we slice our series, the $k$th slice going from $n=k^2$ to $n=(k+1)^2-1$. Then, every term in slice $k$ is at most $e^{-k}$ and slice $k$ has $2k+1$ terms hence

$$\sum_{n=1}^\infty e^{-\sqrt n}\leqslant\sum_{k=1}^\infty (2k+1)e^{-k}\tag{2}$$ 

and all that remains to be shown is that the series in the RHS of $(2)$ converges.
A standard series: Perhaps the most useful series of all is the geometric series, namely the fact that, for every $|x|<1$, 

$$\sum_{k=1}^\infty x^k=\frac1{1-x}\tag{3}$$ 

(Several simple proofs of $(3)$ exist, perhaps you already know some of them.)
Taking this for granted, note that the RHS of $(2)$ is almost a geometric series. To complete the comparison, we differentiate the geometric series term by term on $|x|<1$ (yes, this is legit), yielding  $$\sum_{k=1}^\infty kx^{k-1}=\frac1{(1-x)^2}$$
We shall only keep a small part of this result, namely the fact that the series $$\sum_{k=1}^\infty x^k\qquad\text{and}\qquad\sum_{k=1}^\infty kx^k$$ both converge for every $|x|<1$.
Cauda: In particular, for $x=e^{-1}$, the series in the RHS of $(2)$ converges, qed.
A: 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}}}
%
&= \frac{\exp \left(\, -n\ln \frac{1}{1-\frac{1}{\sqrt{n}}}\,\right)}{e^{-\sqrt{n}}}\\
%
&= \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}}}\\
%
&= 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.
A: 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 $\quad\sum\limits_{k=1}^\infty kx^k=\frac{x}{(1-x)^2}$
A: As you have mentioned $$\exp(-\sqrt n)=\left({1\over e}\right)^{\sqrt n}<{1\over n^2}$$ also for any $0<a<1$ we have $$a=\left({1\over e}\right)^{k}$$where $k=-\ln a>0$ therefore by substitution $$a^{\sqrt n}=\left({1\over e}\right)^{k\sqrt n}=\left({1\over e}\right)^{\sqrt {nk^2}}<{1\over n^2\cdot k^4}$$for large enough $n$. Based on this and on $$\lim_{n\to\infty}\left(1-{1\over \sqrt n}\right)^{\sqrt n}={1\over e}$$we can for small enough $\epsilon>0$ and large enough $n$ write that $$0<{{1\over e}-\epsilon<\left(1-{1\over \sqrt n}\right)^{\sqrt n}<{1\over e}+\epsilon}<1$$therefore $$0<\left({1\over e}-\epsilon\right)^{\sqrt n}<\left(1-{1\over \sqrt n}\right)^{n}<\left({1\over e}+\epsilon\right)^{\sqrt n}<1$$since both $\sum_{n=1}^{\infty}\left({1\over e}+\epsilon\right)^{\sqrt n}$ and $\sum_{n=1}^{\infty}\left({1\over e}-\epsilon\right)^{\sqrt n}$ are convergent then so is $$\sum_{n=1}^{\infty}\left(1-{1\over \sqrt n}\right)^{n}$$
