Let $\displaystyle \mu$ be a measure on the natural numbers s.t $\displaystyle \mu(\left\{n\right\})=\alpha_n$ for all $\displaystyle n\in \mathbb{N}$

Show that a function $\displaystyle f:\mathbb{N}\to\mathbb{R}$ is $\displaystyle \mu$-integrable $\displaystyle \Leftrightarrow

\sum_{n\in\mathbb{N}}|f(n)|\alpha_n $ is convergent.

What is the value of $\displaystyle \int fd\mu $?

I don't quite understand what's to be shown exactly. According to my definitions f is integrable when f is measurable and $\displaystyle \int_{\mathbb{N}}|f|d\mu<\infty$

But isn't the integral equal to $\displaystyle \sum_{n\in \mathbb{N}}|f(n)|d\mu(\left\{n\right\}) = \sum_{n\in \mathbb{N}}|f(n)|\alpha_n....<\infty $?

Isn't $\displaystyle \Rightarrow $ immediate then?

But then what's to be shown for $\displaystyle \Leftarrow $. A little guidance is greatly appreciated