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

Show that a function f:\mathbb{N}\to\mathbb{R} is \mu-integrable  \Leftrightarrow <br /> <br />
\sum_{n\in\mathbb{N}}|f(n)|\alpha_n is convergent.

What is the value of \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 \int_{\mathbb{N}}|f|d\mu<\infty

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

Isn't \Rightarrow immediate then?

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