Show that $\lim_{n\rightarrow \infty}\big[\frac{n^{n+1}+(n+1)^{n}}{n^{n+1}}\big]^{n}=e^{e}$

Question

This is an exercise that a friend of mine asked to me this afternoon.

Show that $\lim_{n\rightarrow \infty}\big[\frac{n^{n+1}+(n+1)^{n}}{n^{n+1}}\big]^{n}=e^{e}.$ All we have done was elementary manipulations, but we got stuck.

I would appreciate any help.

score 6 · Answer 1 · answered Feb 10 '12 at 22:55

6

Using the fact that

$$\lim_{x \to 0} (1+x)^{1/x} = e$$

we can show that if $c_n \to c \neq 0$ then

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

In your case $c_n = (1 + \frac{1}{n})^n$

answered Feb 10 '12 at 22:55

Aryabhata

82,206

This might even be true if $c = 0$, but we don't require it for this problem. – Aryabhata Feb 10 '12 at 22:58
I think you want the $c_n$ to be nonzero... – David Mitra Feb 10 '12 at 23:00
@DavidMitra: No $c \neq 0$ is enough, as for sufficiently large $n$, $|c_n - c| \lt |c|/2$. We don't care if some $c_n$ terms become zero, as ultimately there are all non-zero. – Aryabhata Feb 10 '12 at 23:02
Ah, sorry. Not sure what I was thinking. – David Mitra Feb 10 '12 at 23:03

score 4 · Accepted Answer · answered Feb 10 '12 at 22:55

Define $u_n=(1+\frac{1}{n})^n$.

$$\left(1+\frac{1}{n}\color{Blue}{\left(1+\frac{1}{n}\right)^n}\right)^n $$

$$=\left(1+\frac{1}{n}\color{Blue}{u_n}\right)^n$$

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

$$=\exp\left(u_n+\mathcal{O}\left(\frac{1}{n}\right)\right)$$

The $\mathcal{O}(n^{-1})$ term drops to $0$ and $u_n\to e$ in the limit, so the value sought is $\exp (e) = e^e$.

Show that $\lim_{n\rightarrow \infty}\big[\frac{n^{n+1}+(n+1)^{n}}{n^{n+1}}\big]^{n}=e^{e}$

2 Answers2