How do I show that $\lim_{n \to \infty} (1 + \frac{1}{n+1})^{2n} = e^2$?

Question

It's obvious that $\lim_{n \to \infty} (1 + \frac{1}{n})^{2n} = e^2$? And in the limit, intuitively I can see that the $n+1$ in the denominator of $(1 + \frac{1}{n+1})^{2n}$ shouldn't change the limit. Is there a more explicit method of showing this though or dealing with the $n+1$?

Answer 1

For finite $a,b,c$

$$\left(1+\frac1{an+b}\right)^{cn}=\left[\left(1+\frac1{an+b}\right)^{an+b}\right]^{\dfrac{cn}{an+b}}$$

For $n\to\infty,$ the inner limit converges to $e$

$$\lim_{n\to\infty}\dfrac{cn}{an+b}=\lim_{n\to\infty}\dfrac c{a+b/n}=\dfrac ca$$

Answer 2

Let

$$u_n=\left(1+\frac1n\right)^{2n}$$ so we know that this sequence is convergent to $e^2$ so

$$\left(1+\frac1{n+1}\right)^{2n}=\left(1+\frac1{n+1}\right)^{-2}\left(1+\frac1{n+1}\right)^{2(n+1)}=\left(1+\frac1{n+1}\right)^{-2}u_{n+1}$$ and obviously the subsequence $(u_{n+1})$ is convergent to the same limit $e^2$. Now we conclude using that $$\left(1+\frac1{n+1}\right)^{-2}\xrightarrow{n\to\infty}1$$

Answer 3

You can take $\log$ $$ \log\left[\left(1+\frac{1}{n+1}\right)^{2n}\right]=2n\log\left(1+\frac{1}{n+1}\right)=\frac{2n}{n+1}\cdot\left\{(n+1)\log\left(1+\frac{1}{n+1}\right)\right\}. $$ The term inside $\{\cdot\}$ goes to $1$ whereas $\frac{2n}{n+1}$ goes to $2$.

Answer 4

Another idea :

Make a substitution:

$$m:=n+1\implies \left(1+\frac1{n+1}\right)^{2n}\stackrel{\text{subst.}}\longrightarrow\left(1+\frac1m\right)^{2{m-2}}=$$

$$=\frac{\left[\left(1+\frac1m\right)^m\right]^2}{\left(1+\frac1m\right)^2}\xrightarrow[m\to\infty]{}\frac{e^2}{1^2}=e^2$$

Answer 5

$\lim_{n\to \infty} (1+\frac {1}{n+1})^n$

subst.$ {n+1\to n} $ then $\lim_{n\to \infty} \frac {(1+\frac {1}{n})^{n-1}\cdot (1+\frac {1}{n})}{(1+\frac {1}{n})}=\lim_{n\to \infty} \frac {(1+\frac {1}{n})^{n}}{(1+\frac {1}{n})}=\lim_{n\to \infty} (1+\frac {1}{n})^{n}=e$.