$ \int_{E} |f|d\mu \Leftrightarrow \sum_{n=1}^{\infty}{\frac{1}{n^2} \int_{E}f^21_{\{|f|\leq n\}}d\mu} $ Let $(E,\mathcal {A},\mu)$ be a finite measure space.
Let $f:(E,\mathcal{A}))\to (\mathbb {R},\mathcal{B}(\mathbb{R})) $ be a measurable function.
Show that :
$$
\int_{E} |f|d\mu <\infty \Leftrightarrow \sum_{n=1}^{\infty}{\frac{1}{n^2} \int_{E}f^21_{\{|f|\leq n\}}d\mu} <\infty
$$ 
My first step of demo is :
we have :
$$
\sum_{n=1}^{\infty}{\frac{1}{n^2} \int_{E}f^2 1_{\{|f|\leq n\}}d\mu} =\sum_{n=1}^{\infty}{\frac{1}{n^2} \int_{E}f^2 1_{\biguplus_{k=0}^{n-1}\{k<|f|\leq k+1\}}d\mu} = \sum_{n=1}^{\infty}{\frac{1}{n^2} \sum_{k=0}^{n-1}\int_{E}f^2 1_{\{k<|f|\leq k+1\}}d\mu} = \sum_{k=0}^{\infty}  \left(\sum_{n=k+1}^{\infty}{\frac{1}{n^2}}\right) \left(\int_{E}f^2 1_{\{k<|f|\leq k+1\}}d\mu\right)
$$
but i can't continue, an idea please
 A: There exist constants $c>0$, $C>0$ such that
$$c \frac{1}{k} \leq \sum_{n=k+1}^{\infty} \frac{1}{n^2} \leq C \frac{1}{k}$$
for large $k \gg 1$ (use e.g. integral comparison). By the calculation from your question, this implies
$$\sum_{n=1}^{\infty} \frac{1}{n^2} \int f^2 1_{|f| \leq n} \, d\mu < \infty \iff \sum_{k=1}^{\infty} \frac{1}{k} \int f^2 1_{k<|f| \leq k+1} \, d\mu < \infty. $$
Consequently, it suffices to show
$$\int |f| \, d\mu < \infty \iff \sum_{k=1}^{\infty} \frac{1}{k} \int f^2 1_{k<|f| \leq k+1} \, d\mu < \infty. \tag{1}$$
If $f \in L^1(\mu)$, then
\begin{align*} \sum_{k=1}^{\infty} \frac{1}{k} \int f^2 1_{k<|f| \leq k+1} \, d\mu &\leq \sum_{k=1}^{\infty} \frac{k+1}{k} \int |f| 1_{k<|f| \leq k+1} \, d\mu \\ &\leq 2 \sum_{k=1}^{\infty} \int |f| 1_{k<|f| \leq k+1} \, d\mu \\ &\leq 2 \int |f| \, d\mu < \infty. \end{align*}
On the other hand,
\begin{align*} \int |f| \, d\mu &= \sum_{k=0}^{\infty} \int |f| 1_{k <|f| \leq k+1} \, d\mu \\ &\leq \mu(E) + \sum_{k=1}^{\infty} \int |f| 1_{k <|f| \leq k+1} \, d\mu  \\ &\leq \mu(E) +  \sum_{k=1}^{\infty} \int |f| \frac{|f|}{k} 1_{k <|f| \leq k+1} \, d\mu \\ &= \mu(E) + \sum_{k=1}^{\infty} \frac{1}{k} \int f^2 1_{k<|f| \leq k+1} \ d\mu, \end{align*}
and so
$$\sum_{k=1}^{\infty} \frac{1}{k} \int f^2 1_{k<|f| \leq k+1} \, d\mu < \infty \implies \int |f| \, d\mu < \infty.$$
