$a_n=\int|f|^nd\mu$ then $\lim_{n\to\infty}a_{n+1}/a_n=||f||_\infty$ Let $(X,\mu)$ be a measure space with $\mu(X)=1$. Let $f$ be a measurable function such that $0<||f||_{\infty}< \infty$. Let $a_n=\int|f|^nd\mu$
Show that:
(1) $\lim_{n\to\infty}a_{n+1}/a_n=||f||_\infty$
(2) $\lim_{n\to\infty}a_n^{1/n}=||f||_\infty$
I have no idea how to proceed.

By renormalizing, we can assume $\left\Vert f\right\Vert _{\infty}=1$.
Set $a_{n}:=\int\left|f\right|^{n}\,{\rm d}\mu$ and $b_{n}:=\frac{a_{n+1}}{a_{n}}$
for $n\in\mathbb{N}$. Then $\left\Vert f\right\Vert _{\infty}$ implies
$a_{n+1}\leq a_{n}$ and thus $b_{n}\leq1$ for all $n\in\mathbb{N}$.
Now, the Cauchy-Schwarz inequality implies that
$$
a_{n+1}=\int\left|f\right|^{n+1}\,{\rm d}\mu=\int\left|f\right|^{\frac{n}{2}}\cdot\left|f\right|^{\frac{n+2}{2}}\,{\rm d}\mu\leq\sqrt{\int\left|f\right|^{n}\,{\rm d}\mu}\cdot\sqrt{\int\left|f\right|^{n+2}\,{\rm d}\mu},
$$
i.e. $a_{n+1}^{2}\leq a_{n}\cdot a_{n+2}$. For the quotient $b_{n}=\frac{a_{n+1}}{a_{n}}$,
this implies
$$
b_{n}=\frac{a_{n+1}}{a_{n}}\leq\frac{a_{n+2}}{a_{n+1}}=b_{n+1},
$$
so that $\left(b_{n}\right)_{n\in\mathbb{N}}$ is nondecreasing and
bounded above by $1$. Thus, $b:=\lim_{n\to\infty}b_{n}=\sup_{n\in\mathbb{N}}b_{n}\in\left(0,1\right]$
exists.
Now, the ratio test easily implies that the power series
$$
\sum_{n=1}^{\infty}a_{n}x^{n}
$$
has radius of convergence $R=\frac{1}{b}$.
But as noted in the comments, the question Limit of $L^p$ norm
shows that we have
$$
\sqrt[n]{a_{n}}\xrightarrow[n\to\infty]{}\left\Vert f\right\Vert _{\infty}=1.
$$
Using the root test, we see that the power series from above also
has the radius of convergence $R=\frac{1}{\lim_{n\to\infty}\sqrt[n]{a_{n}}}=1$.
All in all, we conclude $\lim_{n\to\infty}b_{n}=R=1=\left\Vert f\right\Vert _{\infty}$,
as desired. 
EDIT: What I am using here is in more detail as follows: From above, we know that the limit $b := \lim_{n \to \infty} \frac{a_{n+1}}{a_n}$ exists.
Now, if $|x| < 1/b$, we have
$$
\bigg|\frac{a_{n+1}x^{n+1}}{a_n x^n}\bigg|  = \frac{a_{n+1}}{a_n} \cdot |x| \to b |x| <1,
$$
so that the series $\sum_n a_n x^n$ converges (absolutely).
Conversely, if $|x|> 1/b$, a similar argument shows that the series diverges. Hence, the radius of convergence is given by $1/b$.
Essentially the same holds for the root test.
