For which $p$ do we have $\lim_{n \to \infty}\left(1 + \frac1{n^{p}} \right)^{n^p} = e$?

We all known that $$ \lim_{n \to \infty}\; ( 1 + 1/n )^n = {\rm e} \; . $$ My question is, for which $p$ does the limit still be ${\rm e}$, when $n$ is replaced by $n^p$ ? $$ \lim_{n \to \infty} \;( 1 + 1/n^{p} )^{n^{p}} \;?=?\; {\rm e} \; . $$ For example, when $p=\frac{1}{2}$, then $$ \lim_{n \to \infty} \; \left( 1 + 1/\sqrt{n} \right)^{\sqrt{n}} = {\rm e} \; . $$ But when $p=-\frac{1}{2}$, $$ \lim_{n \to \infty} \; \left[ \left( 1+\sqrt{n} \right) ^{\frac{1}{\sqrt{n}}} \right] = 1 \;. $$

Over all $p \in \mathbb{R}$, what are the possibilites for those limits?

Solution 1:

A simple substitution of $u = n^p$ gives that

$$L := \lim_{n \to \infty} \left( 1 + \frac{1}{n^p} \right)^{n^p} = \lim_{u \to c} \left( 1 + \frac{1}{u} \right)^u$$

The question being, what is this "$c$" term I introduced? More clearly,

$$c = \lim_{n \to \infty} n^p$$

If $L = e$, then we need $c = \infty$. For that to happen we need $p \in (0,\infty)$.

(If $p = 0$ then $c=1$ and $L=2$; if $p<0$, then $c = 0$ and $L=1$.)

Hence, we conclude that $p > 0$.

Solution 2:

More generally, $\lim_{h\to 0} (1+h)^{\frac{1}{h}} = e$. So whenever $p>0$ we have $\lim_{n\to\infty} (1+\frac{1}{n^p})^{n^p} = e$.

For $p=0$ we get the constant sequence $2$, and when $p<0$ we have $$\frac{1}{n^p}<1+\frac{1}{n^p} < \frac{2}{n^p}$$ so by the squeeze theorem $(1+\frac{1}{n^p})^{n^p}\to 1$.

(To elaborate on the last case: $(\frac{1}{n^p})^{n^p}=e^{-n^p \ln(n^p)}$, and that $n^p \ln(n^p)\to 0$ can be shown by L'Hopital's rule).