Ytukyg

I believe there are several ways to prove that $sumlimits^{infty}_{n=1}left(1-frac{1}{sqrt{n}}right)^n$ converges. Can you, please, post yours so that we can learn from you?

HERE IS ONE

Let $ninBbb{N}$ 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 \&=explnleft(1-frac{1}{sqrt{n}}right)^n\&=exp left[nlnleft(1-frac{1}{sqrt{n}}right)right] \&=expleft[ -nsum^{infty}_{k=1}frac{1}{k}left(frac{1}{sqrt{n}}right)^kright]\&=expleft[ -nleft(frac{1}{sqrt{n}}+frac{1}{2n}+sum^{infty}_{k=3}frac{1}{k}left(frac{1}{sqrt{n}}right)^kright)right]\&=exp left[-sqrt{n}-frac{1}{2}-sum^{infty}_{k=3}frac{n}{k}left(frac{1}{sqrt{n}}right)^kright]\&equivexp 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}toexp left(-frac{1}{2}right).end{align}
Since $b_n to 0$, there exists $N$ such that for all $ngeq N,$
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, $sum^{infty}_{n=1}b_n<inftyimplies sum^{infty}_{n=1}a_n<infty$ by Limit comparison test.

As you have mentioned $$exp(-sqrt n)=left({1over e}right)^{sqrt n}<{1over n^2}$$ also for any $0<a<1$ we have $$a=left({1over e}right)^{k}$$where $k=-ln a>0$ therefore by substitution $$a^{sqrt n}=left({1over e}right)^{ksqrt n}=left({1over e}right)^{sqrt {nk^2}}<{1over n^2cdot k^4}$$for large enough $n$. Based on this and on $$lim_{ntoinfty}left(1-{1over sqrt n}right)^{sqrt n}={1over e}$$we can for small enough $epsilon>0$ and large enough $n$ write that $$0<{{1over e}-epsilon<left(1-{1over sqrt n}right)^{sqrt n}<{1over e}+epsilon}<1$$therefore $$0<left({1over e}-epsilonright)^{sqrt n}<left(1-{1over sqrt n}right)^{n}<left({1over e}+epsilonright)^{sqrt n}<1$$since both $sum_{n=1}^{infty}left({1over e}+epsilonright)^{sqrt n}$ and $sum_{n=1}^{infty}left({1over e}-epsilonright)^{sqrt n}$ are convergent then so is $$sum_{n=1}^{infty}left(1-{1over sqrt n}right)^{n}$$

$$
frac{left(,1-frac{1}{sqrt{n}},right)^n}{e^{-sqrt{n}}}
= frac{left(,left(,1-frac{1}{sqrt{n}},right)^{sqrt{n}},right)^{sqrt{n}}}{left(,e^{-1},right)^{sqrt{n}}} to 1
$$
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.

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 $x$,

$$1-xleqslant e^{-x}tag{1}$$

(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 $xleqslant1$ and every nonnegative $n$, $$(1-x)^nleqslant e^{-nx}$$ for example, for every positive $n$, $$left(1-frac1{sqrt n}right)^nleqslant e^{-sqrt n}$$ 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

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

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 $$sum_{k=1}^infty x^kqquadtext{and}qquadsum_{k=1}^infty kx^k$$ both converge for every $|x|<1$.

Cauda: In particular, for $x=e^{-1}$, the series in the RHS of $(2)$ converges, qed.