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I have trouble evaluating this limit: $$\lim_{n \to \infty} \left ( \sqrt[n]{n} + \frac 1 n \right )^{\frac n {\ln n}}$$

I cannot use series expansion... I also tried to rewrite it as $\large e^{\ln (\cdots)}$, without getting anywhere. It's the $\sqrt[n]{n}$ factor that troubles me the most. I can't get rid of it.

rubik
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  • "I cannot use series expansion..." One can very much. – Did Sep 12 '14 at 19:46
  • @Did: I suspected there were some ways I didn't see. – rubik Sep 12 '14 at 19:49
  • Thus, show what you tried. – Did Sep 12 '14 at 19:50
  • Ah I think I have it. Rewriting the expression as $e^{\frac{n}{\ln n} \ln \left ( n^{\frac 1n} + \frac 1n \right )}$ may be the right first step. $\frac 1n$ tends to $0$, and $n^{\frac 1n}$ tends to $1$, so I can expand the log? – rubik Sep 12 '14 at 19:57
  • I know that $\ln(1 + \epsilon(x)) \sim \epsilon(x) + o(\epsilon(x))$ if $\epsilon(x) \to 0$. But does it apply even if $\ln(\delta(x) + \epsilon(x))$, with $\delta(x) \to 1$? I think not, because I get a result of $0$, while the book says $e$. – rubik Sep 12 '14 at 20:01
  • No, when $\delta(x)\to1$ and $\epsilon(x)\to0$, $\log(\delta(x)+\epsilon(x))=\delta(x)+\epsilon(x)-1+o(\delta(x)+\epsilon(x)-1)$. – Did Sep 12 '14 at 20:11

3 Answers3

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Factor out $n^{1/n}$ via $\log(n^{1/n}+n)=\log(n^{1/n})+\log(1+1/n^{1+1/n})$ and simplify. You can taylor expand the second $\log$ if necessary. It looks the answer will be $e$.

Alex R.
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2

With no trick, using only that $\mathrm e^x=1+x+o(x)$ and $\log(1+x)=x+o(x)$ when $x\to0$.

(Note that these two expansions are actually equivalent since $\exp$ and $\log$ are mutually inverse.)

  • $\sqrt[n]{n}=\mathrm e^{\log n/n}=1+\log n/n+o(\log n/n)$
  • $1/n=o(\log n/n)$
  • $\sqrt[n]{n}+1/n=1+\log n/n+o(\log n/n)$
  • $\log(\sqrt[n]{n}+1/n)=\log n/n+o(\log n/n)$
  • $(n/\log n)\log(\sqrt[n]{n}+1/n)=1+o(1)$
  • $\exp(1+o(1))\to\mathrm e$
Did
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2

Use the fact that if $a_n \rightarrow a$ then $$\left(1+\frac{a_n}{n}\right)^n \rightarrow e^a$$

We write our expression as

$$\left(\sqrt[n]{n}+\frac{1}{n}\right)^{\frac{n}{\ln n}}=\left(1+\frac{n(\sqrt[n]{n}-1)+1}{n}\right)^{\frac{n}{\ln n}}= \left(1+\frac{a_n}{n/\ln n}\right)^{\frac{n}{\ln n}}$$ where $$a_n=\frac{n(\sqrt[n]{n}-1)+1}{\ln n}$$ so it is reduced to finding this last limit. $\frac{1}{\ln n}\rightarrow 0$ so we want the limit of $\frac{n(\sqrt[n]{n}-1)}{\ln n}$

But this follows immediately from $$\lim\limits_{x\rightarrow 0} \frac{e^x-1}{x}=1$$ let $x=\frac{\ln n}{n}$ to get

$$\lim\limits_{n\rightarrow \infty } \frac{\sqrt[n]{n}-1}{\ln n/n}=1$$ therefore $a_n \rightarrow 1$ and our limit for the original expression will be $e$. In agreement with the other answers.

Rene Schipperus
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