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Knowing that the number $e =\lim_{n\to\infty}\left(1+{1\over{n}}\right)^n$ solve $a_n=\left({n+1}\over{n+3}\right)^n$

So (...) $$\lim_{n\to\infty}\left({n+1}\over{n+3}\right)^n = \lim_{n\to\infty}\left({n+1+2-2}\over{n+3}\right)^n = \lim_{n\to\infty}\left(1-{2\over{n+3}}\right)^n = \lim_{n\to\infty}\left({1+{{1\over{-n-3}}\over2}}\right)^n = \lim_{n\to\infty}\left({1+{{1\over{-n-3}}\over2}}\right)^{{-n-3\over2}.{2\over{-n-3}}.n} = e^{\lim_{n\to\infty}{2n\over{-n-3}}}=e^{-2}$$

I know to solve it in this way, but my teacher told me that the in exams I can NOT solve it in this way. The problem is that I can not put the denominator in the exponent and then multiply by the inverse to make it 1 (note that this does not alter the exercise).

Is there another way to do this?

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    $\begingroup$ The general approach you have is great; the only problem is that $\lim_{n\rightarrow \infty}\frac{-n-3}{2}=-\infty$, not $+\infty$. For this reason you can't use the $e$ result. $\endgroup$ – vadim123 May 9 '13 at 1:41
  • $\begingroup$ Thanks, but this is evident, because the n is negative, when I replace n with ∞, this will be −∞ $\endgroup$ – Tomi May 9 '13 at 1:44
  • $\begingroup$ @vadim123 Well, one can alternatively use that to prove that $$\left(1-\frac 1 n\right)^n\to e^{-1}$$ and move on! $\endgroup$ – Pedro Tamaroff May 9 '13 at 1:44
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Maybe go in this direction instead: $$\left(1-\frac{2}{n+3}\right)^n=\left(1-\frac{2}{n+3}\right)^{n+3-3}=\left(1-\frac{2}{n+3}\right)^{n+3}\left(1-\frac{2}{n+3}\right)^{-3}$$The right term goes to $1$ as $n\rightarrow \infty$, and the left term goes to $e^{-2}$.

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yes there is, we have$$\left(\frac{n+1}{n+3}\right)^n=\left(\frac{\frac{n+1}{n}}{\frac{n+3}{n}}\right)^n=\left(\frac{1+\frac{1}{n}}{1+ \frac{3}{n}}\right)^n=\frac{(1+\frac{1}{n})^n}{(1+\frac{3}{n})^n}\longrightarrow\frac{e}{e^3}=e^{-2}.$$and we know the limits of the RHS.

Edit: you can show that $(1+\frac{a}{n})^n\to e^a$ and use this fact to calculate the limit you want.

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Look at the reciprocal. This is $$\left(\frac{n+3}{n+1}\right)^n=\left(1+\frac{2}{n+1}\right)^n.\tag{$1$}$$ Now we make a little mistake. Rewrite this as $$\left(\left(1+\frac{1}{(n+1)/2}\right)^{(n+1)/2}\right)^{2n/(n+1)}.$$ Let $n\to\infty$. Then $\left(1+\frac{1}{(n+1)/2}\right)^{(n+1)/2}\to e$ and $\frac{2n}{n+1}\to 2$, so our expression $(1)$ has limit $e^2$. It follows that the original expression has limit $\frac{1}{e^2}$.

The mistake: For $n$ odd, everything is fine, since $\frac{n+1}{2}$ is an integer. For $n$ even, there is a technical problem, since $\frac{2}{n+1}$ is not an integer. We can fix things for even $n$ by showing that the limit of $$\lim_{n\to\infty}\left(1+\frac{2}{n}\right)^n=e^2\quad\text{and}\quad\lim_{n\to\infty}\left(1+\frac{2}{n+2}\right)^n=e^2,$$ and the result follows by squeezing.

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