Spivak, Calculus, Ch. 22: How do we compute the limit $\lim\limits_{n\to\infty} \frac{(n+1)^{\frac{n+1}{n}}}{n}$? My question is simply how do we compute the limit
$$\lim\limits_{n\to\infty} \frac{(n+1)^{\frac{n+1}{n}}}{n}$$
I know the limit is $1$, both from the context below and because I checked in Maple.
Here is the context in which this limit arose.
In Chapter 22 of Spivak's Calculus, Problem 13 asks us to show first that if $f$ is increasing on $[1,\infty)$ then
$$f(1)+...+f(n-1)<\int_1^n f(x)dx<f(2)+...+f(n)$$
When we apply this result to the function $f(x)=\log{(x)}$ we easily obtain the relationship
$$\frac{n^n}{e^{n-1}}<n!<\frac{(n+1)^{n+1}}{e^n}\tag{1}$$
Spivak concludes for us that given this result we can say that
$$\lim\limits_{n\to\infty} \frac{\sqrt[n]{n!}}{n}=\frac{1}{e}\tag{2}$$
I am interested in going from $(1)$ to $(2)$.
Starting at $(1)$, if we take the n-th root and divide by $n$ we have
$$0<\frac{1}{e^{\frac{n-1}{n}}}<\frac{\sqrt[n]{n!}}{n}<\frac{(n+1)^{\frac{n+1}{n}}}{n}\cdot \frac{1}{e}$$
Now,
$$\lim\limits_{n\to\infty} \frac{1}{e^{\frac{n-1}{n}}}=\frac{1}{e}$$
I would like to compute the limit
$$\lim\limits_{n\to\infty} \frac{(n+1)^{\frac{n+1}{n}}}{n}\cdot \frac{1}{e}$$
Which necessitates computing the limit that gave rise to the current question
$$\lim\limits_{n\to\infty} \frac{(n+1)^{\frac{n+1}{n}}}{n}$$
 A: I would start with noticing that
\begin{align*}
\frac{(1 + n)^{\frac{n+1}{n}}}{n} & = \frac{(1 + n)^{1 + \frac{1}{n}}}{n} = \left(1 + \frac{1}{n}\right)\times(1 + n)^{1/n}
\end{align*}
where the last limit can be computed as follows:
\begin{align*}
\lim_{n\to\infty}(1 + n)^{1/n} = \lim_{n\to\infty}\exp\left(\frac{\ln(1 + n)}{n}\right) = \exp\left(\lim_{n\to\infty}\frac{\ln(1 + n)}{n}\right) & = \exp(0) = 1
\end{align*}
Hence one concludes that
\begin{align*}
\lim_{n\to\infty}\frac{(1 + n)^{\frac{n+1}{n}}}{n} = \lim_{n\to\infty}\left(1 + \frac{1}{n}\right)\times(1 + n)^{1/n} = 1\times 1 = 1
\end{align*}
Hopefully this helps!
A: Another way:
Using $$ \lim_{n \to \infty} \frac{1}{n} \, \ln\left(1 + \frac{1}{n}\right) = 0$$ and $\frac{\ln n}{n} < 1$ then
$$ \frac{1}{n} \, (n+1)^{\frac{n+1}{n}} = \frac{n+1}{n} \, e^{\left( \ln n + \ln\left(1 + \frac{1}{n}\right) \right)/n} = \left(1 + \frac{1}{n}\right) \, e^{\frac{1}{n} \, \ln\left(1 + \frac{1}{n}\right)} \, e^{\ln n/n} $$
which gives the limit
\begin{align}
L &= \lim_{n \to \infty} \, \frac{1}{n} \, (n+1)^{\frac{n+1}{n}} \\
&= \lim_{n \to \infty} \,  \left(1 + \frac{1}{n}\right) \, e^{\frac{1}{n} \, \ln\left(1 + \frac{1}{n}\right)} \, e^{\ln n/n} \\
&= \lim_{n \to \infty} \,  \left(1 + \frac{1}{n}\right) \, e^{\frac{1}{n} \, \ln\left(1 + \frac{1}{n}\right)} \, \left(1 + \frac{\ln n}{n} + \frac{\ln^2 n}{2 \, n^2} + \mathcal{O}\left(\frac{\ln^3 n}{n^3}\right) \right) \\
&= 1
\end{align}
A: $$L=\lim\limits_{n\to\infty} \frac{(n+1)^{\frac{n+1}{n}}}{n}$$
$$L=\lim_{x\to \infty}\frac{n+1}{n} \lim_{n\to \infty}(1+n)^{1/n}=1.\exp[ \lim_{n \to \infty}\frac{\ln(1+n)}{n}]=\exp[ \lim_{n\to \infty} \frac{1}{1+n}]=e^0=1$$.
Last part is by L'Hosp.
