Determine if a sequence converges using the number e 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?
 A: 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 $e^{-2}$.
A: 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.
A: 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. 
