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This is what I've done:

$$\frac{n^n}{(n+1)^n} = \left(\frac{n}{n+1}\right)^n = \exp(\log\left(\left(\frac{n}{n+1}\right)^n\right) = \exp(n\log\left(\frac{n}{n+1}\right)$$

And then I can take the limit like this:

$$\exp\left(\lim\limits_{n \to \infty}n\log\left(\frac{n}{n+1}\right)\right)$$

But where to from here?

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  • $\begingroup$ Hint: what is the definition of $e$ ? $\endgroup$ – user223391 Sep 23 '15 at 1:38
  • $\begingroup$ The limit of the reciprocal may be already familiar. $\endgroup$ – André Nicolas Sep 23 '15 at 1:38
  • $\begingroup$ How about $\frac{lnn-ln(n+1)}{1/n}$ and use L'Hospital Rule? $\endgroup$ – imranfat Sep 23 '15 at 1:39
  • $\begingroup$ @AndréNicolas. You are right but perhaps he has to prove the limit as an exercise... $\endgroup$ – imranfat Sep 23 '15 at 1:40
  • $\begingroup$ As Andre' says, just divide numerator and denominator by $n^n$ and the limit in the reciprocal is familiar. Usually school courses require you to show it that way. $\endgroup$ – Shailesh Sep 23 '15 at 1:46
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Without resorting to L'Hopital and using only the basic equalities, \begin{align*} \lim_{n \to \infty} \frac{(n+1)^n}{n^n} &= \lim_{n \to \infty} \left(\frac{n+1}{n}\right)^n \\ &= \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \\ &= e, \end{align*} so the reciprocal tends to $1/e$.

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  • $\begingroup$ Selecting this as best answer since the way I approached the problem was quite contrived, and this shows the simpler and better way to do it. As other commenters noted, the expression $\frac{n^n}{(n+1)^n}$ simplifies to $\frac{1}{(n+\frac{1}{n})^n}$ and taking the limit of this gives $\frac{1}{e}$ $\endgroup$ – PythonNewb Sep 23 '15 at 2:03
  • $\begingroup$ I wouldn't be so quick to dismiss your approach. It's a standard technique to turn $a^x$ into $e^{x \ln a}$ $\endgroup$ – user217285 Sep 23 '15 at 2:10
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$$\exp\left(\lim\limits_{n \to \infty}n\log\left(\frac{n}{n+1}\right)\right)$$ $$= \exp\left(\lim\limits_{n \to \infty}\frac{\log\left(\frac{n}{n+1}\right)}{\frac{1}{n}}\right)$$ $$= \exp\left(\lim\limits_{n \to \infty}\frac{\frac{1}{n^2 + n}}{\frac{-1}{n^2}}\right)$$ $$= \exp\left(\lim\limits_{n \to \infty}\frac{-n^2}{n^2 + 1}\right)$$ $$= \exp(-1)$$ $$ = \frac{1}{e}$$

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If you accept that $e$ is the unique number such that $1 = ∫_1^e \frac{dt}{t}$, then there is a proof here of $(1+\frac{1}{n})^n→ e$ which then gives by Nitin's argument the result.

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