Convergence of $\sum\limits_{n=1}^{\infty}\left(\frac{\sqrt n-1}{\sqrt n}\right)^n$

Question

$$\sum_{n=1}^{\infty}\left(\frac{\sqrt n-1}{\sqrt n}\right)^n$$ It kind of looks like the Euler's number limit, but I didn't succeed in proving that it converges. Anyone? This was an exercise at the end of a chapter on sequences and series, much before Taylor series and integrals, so I don't think it would be "fair" to use those.

Answer 1

Note that $1-x\leqslant\mathrm e^{-x}$ for every $x$. Applying this to $x=1/\sqrt{n}$ yields $$ \left(\frac{\sqrt{n}-1}{\sqrt{n}}\right)^n\leqslant\left(\mathrm e^{-1/\sqrt{n}}\right)^n=\mathrm e^{-\sqrt{n}}, $$ hence it suffices to show that the series $$ \sum_{n\geqslant1}\mathrm e^{-\sqrt{n}} $$ converges. There are several ways to do so, an instructive one is to note that the function $x\mapsto\mathrm e^{-\sqrt{x}}$ is decreasing hence, for every $n\geqslant1$, $$ \mathrm e^{-\sqrt{n}}\leqslant\int_{n-1}^n\mathrm e^{-\sqrt{x}}\mathrm dx. $$ Summing these yields $$ \sum_{n\geqslant1}\mathrm e^{-\sqrt{n}}\leqslant I,\qquad\text{with}\quad I=\int_0^\infty\mathrm e^{-\sqrt{x}}\mathrm dx. $$ Now, the change of variable $x=t^2$ yields $$ I=2\int_0^\infty t\mathrm e^{-t}\mathrm dt=2\,\left.(t+1)\mathrm e^{-t}\right|_0^\infty=2, $$ which is finite, hence the series of interest indeed converges.

Another approach, with no integral, is to show that there exists some finite $c$ such that, for every $n$, $$ \mathrm e^{-\sqrt{n}}\leqslant\frac{c}{n^2}, $$ since every series with general term $c/n^2$ converges.

Starting from the elementary fact that $\mathrm e^x\geqslant x$ for every $x$, one gets successively $\mathrm e^{x/4}\geqslant x/4$, then $\mathrm e^{x}\geqslant x^4/4^4$ for every $x\geqslant0$, which, for $x=\sqrt{n}$, yields $$ \mathrm e^{-\sqrt{n}}\leqslant\frac{256}{n^2}. $$ To sum up, the argument with no integral is that, for every $n\geqslant1$, $$ \left(\frac{\sqrt{n}-1}{\sqrt{n}}\right)^n\leqslant\frac{256}{n^2}. $$

Answer 2

If $f(n) = \left(1 - n^{-1/2}\right)^n$, then $$\ln(f(n)) = n \ln(1 - n^{-1/2}) = n (-n^{-1/2} + O(n^{-1})) = - n^{1/2} + O(1) < -2 \ln(n)$$ for $n$ sufficiently large. Thus $f(n) < n^{-2}$ for such $n$.

Answer 3

Well, $$ \left(\frac{\sqrt{n}-1}{\sqrt{n}}\right)^n = \left[\left(1-\frac{1}{\sqrt{n}}\right)^{\sqrt{n}}\right]^{\sqrt{n}} \to \mathrm{e}^{-\sqrt{n}}. $$

This suggests a limit comparison test to $\sum_n \mathrm{e}^{-\sqrt{n}}$; one way to determine this "simpler" series convergence is by the integral test.

Answer 4

$e^{- \sqrt{n}} \geq \left( \frac{ \sqrt{n} - 1}{\sqrt n} \right)^n = \left((1 - \frac{1}{n^{1/2}})^{n^{1/2}}\right)^{n^{1/2}} \quad \forall n \geq 1$ (since the RHS approaches the LHS from below).

Here is a method to show that $\displaystyle \sum_{n=1}^{\infty} e^{- \sqrt{n}}$ converges, so by comparison your series converges as well.