Several ways to prove that $\sum\limits^\infty_{n=1}\left(1-\frac1{\sqrt{n}}\right)^n$ converges

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $\quad\sum\limits_{k=1}^\infty kx^k=\frac{x}{(1-x)^2}$

Answer 4

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 $$\sum_{n=1}^{\infty}\left(1-{1\over \sqrt n}\right)^{n}$$