Proof that $E[X^2]$ = $\sum_{n=1}^\infty (2n-1) P(X\ge n)$

X is a random variable with values from $$\Bbb N\setminus{0}$$

I am trying to show that $$E[X^2]$$ = $$\sum_{n=1}^\infty (2n-1) P(X\ge n)$$ iff $$E[X^2]$$ < $$\infty$$.

I rewrote $$P(X \ge n)$$:

$$E[X^2]$$ = $$\sum_{n=1}^\infty (2n-1)\sum_{x=1}^\infty 1_{x \ge n}P(X=x)$$

Now I tried to rearrange the sums:

$$E[X^2]$$ = $$\sum_{x=1}^\infty \sum_{n=1}^x (2n-1)P(X=x)$$

But I think that I made a mistake. Could you give me some hints?