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

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

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 $n\geqslant1$, $$\left(\frac{\sqrt{n}-1}{\sqrt{n}}\right)^n\leqslant\frac{256}{n^2}.$$

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 $n$.

• this was an exercise at the end of a chapter on succession and series, and before taylor series and o notation. Do you think it is possible to prove that it converges using only the comparison test? Apr 6, 2014 at 18:50

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.

• I should add that my limit-taking process involved lots of hand-waving. How much rigor do you need in the limit? Apr 6, 2014 at 19:11
• The point is that this was a chapter-end exercise on sequences and series, and much before integrals, so I cannot use that test. The only test that it would be fair to use is the comparison test. Apr 6, 2014 at 19:15
• "Don't waste your time seeking ways with more rigor. Just raise up your hands and wave with more vigor!" =) Apr 6, 2014 at 19:20
• @DavidH, that's a great quote! Do you know the source? (Is it you?) Apr 6, 2014 at 20:28
• @AntonioVargas I have no idea, but I'll just confidently assert it was Richard Feynman and wave my hands vigorously. Apr 6, 2014 at 20:36

$e^{- \sqrt{n}} \geq \left( \frac{ \sqrt{n} - 1}{\sqrt n} \right)^n = \left((1 - \frac{1}{n^{1/2}})^{n^{1/2}}\right)^{n^{1/2}} \quad \forall n \geq 1$ (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.