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I have the following limit

$$\lim_{n\rightarrow+\infty}{\left({\frac{n}{n+1}}\right)^n}$$

It leads to an indeterminate $[1^\infty]$, this means I should look into the limit of $e$. I rework the function a little bit:

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

This is similar but not identical to

$$\left({1+\frac{1}{n}}\right)^n\rightarrow e$$

The major difference I notice is the $n+1$ in the denominator. Is the limit considered to approach $e^{-1}$ because for $n\rightarrow\infty$ we have $n+1\sim n$?

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  • $\begingroup$ Well, you could multiply (and divide) the equation with $(1-1/(n+1))$. Then you can use the known limit for the first part and for the second part you just use that the limit is given by $1$. $\endgroup$ – Stan Tendijck Sep 4 '18 at 7:58
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Remark that

$$\left(\frac n{n+1}\right)^n=\left(1+\frac1n\right)^{-n}$$ and by continuity of the reciprocal function, the limit of the reciprocal is the reciprocal of the limit.


This shows that in general $n+1\sim n$, unless there is some cancellation. (More specifically, $n+1-n\nsim n-n$.)

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You can write $$\left(\frac{n}{n+1}\right)^n=\left(\left(1-\frac1{n+1}\right)^{n+1}\right)^{\frac{n}{n+1}}=a_n^{\frac{n}{n+1}}$$where $\lim_{n\to\infty}a_n=e^{-1}$ and $\lim_{n\to\infty}\frac{n}{n+1}=1$.

Then: $$\lim_{n\to\infty}a_n^{\frac{n}{n+1}}=\left(e^{-1}\right)^1=e^{-1}$$

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  • $\begingroup$ I love this method. (+) $\endgroup$ – Math Sep 4 '18 at 8:40
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Note that: $$\lim_{n\to\infty}\left(1+\frac1n\right)^n=\lim_{n\to-\infty}\left(1+\frac1n\right)^n=\lim_{n\to+\infty}\left(1+\frac1n\right)^n=e.$$ So, one way: $$\lim_{n\rightarrow+\infty}{\left({\frac{n}{n+1}}\right)^n}=\lim_{n\rightarrow+\infty}{\left({\frac{1}{\frac{n+1}{n}}}\right)^n}=\\ \lim_{n\rightarrow+\infty}{\frac{1}{\left(1+\frac{1}{n}\right)^n}}=\frac{1}{\lim_\limits{n\rightarrow+\infty}\left(1+\frac{1}{n}\right)^n}=\frac1e.$$ or the other way: $$\lim_{n\to+\infty}\left({1-\frac{1}{n+1}}\right)^n=\lim_{n\to+\infty}\left(\left[{1+\frac{1}{-(n+1)}}\right]^{-(n+1)}\right)^{\frac{n}{-(n+1)}}=\\ e^{\lim_\limits{n\to+\infty} \frac{n}{-(n+1)}}=e^{-1}=\frac1e.$$

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$$L=\lim_{n\rightarrow+\infty}{\left({\frac{n}{n+1}}\right)^n}$$ $$\log L=\lim_{n\rightarrow+\infty}\log {\left({\frac{n}{n+1}}\right)^n}$$ $$\log L=\lim_{n\rightarrow+\infty}n\log {\left({\frac{n}{n+1}}\right)}$$ $$\log L=\lim_{n\rightarrow+\infty}\frac{\log {\left({\frac{n}{n+1}}\right)}}{\frac{1}{n}}$$

Apply LHopital's rule to the RHS

$$\log L=\lim_{n\rightarrow+\infty}\frac{\frac{1}{n^2+n}}{\frac{-1}{n^2}}$$ $$\log L=-1\lim_{n\rightarrow+\infty}\frac{n^2}{n^2+n}=-1$$

$$L=e^{-1}$$

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Recall that $\forall a\in \mathbb{R}$

$$\left({1+\frac{1}{n}}\right)^n\rightarrow e\implies \left({1+\frac{a}{n}}\right)^n\rightarrow e^a$$

therefore from your step

$$\left({\frac{n}{n+1}}\right)^n=\left({\frac{n+1-1}{n+1}}\right)^n=\left({1-\frac{1}{n+1}}\right)^n=\left({1+\frac{-1}{n+1}}\right)^{n}=\frac{\left({1+\frac{-1}{n+1}}\right)^{n+1}}{\left({1+\frac{-1}{n+1}}\right)}\to \frac{e^{-1}}{1}=e^{-1}$$

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