I understand that this may require the use of the more popular limit:

$$\lim_{n \to \infty} (1 + \frac{a}{n})^n = e^a$$

but can't seem to find the correct way to use it.

I've tried setting $b_n$ := $(n^{1/n} - 1)$ so that $(1 + b_n)^n = n \to \infty$, but that didn't seem to help.

Any hints on this method or solutions using other methods would be greatly appreciated. Thanks.

  • 2
    $\begingroup$ Have you tried to take logarithims of both sides, put it in $0/0$ or $\infty/\infty$ form then apply Lopital's rule? $\endgroup$ – Baby Dragon Nov 10 '12 at 17:10
  • 1
    $\begingroup$ I suspect that "Lopital" should be written "l'Hôpital"! :) $\endgroup$ – Giovanni De Gaetano Nov 10 '12 at 17:16
  • $\begingroup$ @GiovanniDeGaetano: That's how we write it, but it is my understanding that being obsessed with how one spells names is a modern phenomenon. l'Hôpital himself may possibly have used a variety of spellings. But “Lopital” may not have been among them. $\endgroup$ – Harald Hanche-Olsen Nov 10 '12 at 17:26
  • $\begingroup$ @HaraldHanche-Olsen You mention a fact I wasn't aware of, but which is certainly true in the spirit! For example Wikipedia mention the second possible spelling: "L'Hospital". Thank you very much for that! :) Regarding the specific spelling "Lopital" I doubt it was used, essentially for etymological reasons, but it doesn't look like a bad error anymore! $\endgroup$ – Giovanni De Gaetano Nov 10 '12 at 17:33

You want $n(n^{\frac{1}{n}}-1)>B$, or $n^{\frac{1}{n}}-1>B/n$, or $n>(1+B/n)^n$, which is true for large $n$ as the left hand side goes to infinity and right hand side goes to $e^B$.


Using power series we get $$ \frac1x\left(e^x-1\right)=1+\frac12x+\frac16x^3+\dots $$ thus, setting $x=\frac{\log(n)}{n}$ yields $$ \frac{n}{\log(n)}\left(e^{\frac1n\log(n)}-1\right) =1+\frac12\left(\frac1n\log(n)\right)+\frac16\left(\frac1n\log(n)\right)^3+\dots $$ Therefore, $$ \lim_{n\to\infty}\frac{n}{\log(n)}\left(n^{1/n}-1\right)=1 $$ and $$ n\left(n^{1/n}-1\right)\sim\log(n) $$ so that $$ \lim_{n\to\infty}n\left(n^{1/n}-1\right)=\infty $$

  • $\begingroup$ Why does $n\left(n^{1/n}-1\right)\sim\log(n)$? $\endgroup$ – Joe Nov 10 '12 at 20:31
  • $\begingroup$ @Joe: that is simply another way of saying $$\lim_{n\to\infty}\frac{n}{\log(n)}\left(n^{1/n}-1\right)=1$$ $\endgroup$ – robjohn Nov 10 '12 at 22:24
  • $\begingroup$ Oh, wow. Simply glossed over that when I was tired earlier. I thought you had wrote $\left(n^{1/n}-1\right)\sim\log(n)$. $\endgroup$ – Joe Nov 10 '12 at 23:55

Here is a way: Define $f_n(x)=n^x$ and use the mean value theorem; $$n^{1/n}-1=f_n(1/n)-f_n(0)=\frac1nf_n'(c_n)$$ for some $c_n$ between $0$ and $1/n$. I hope you can compute $f_n'(x)$ and then take it the rest of the way yourself?


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