probability - f(x)=(1/2)^x, x=1,2,3,4, ...find the mean

A fair coin is flipped successively at random until the first head is observed. Let the random variable X denote the number of flips of the coin that are required. Then the space of x is S={x: x=1,2,3,4, ....}. Later we learn that, under certain conditions, we can assign probabilities to these outcomes in S with the function $f(x)=(1/2)^x, x=1,2,3,4,...$ Compute the mean $\mu$.

I know $\mu =E(X)= \sum_{x=1}^ \infty x \cdot f(x)=1 \cdot (1/2)^1+2 \cdot(1/2)^2+3\cdot (1/2)^3+....=1/2+1/2+3/8+...$

I have thought about factoring 1/2 out, but I still could not figure out the mean. I know $\sum_{x=1} ^ \infty f(x)=1.$ I just need the help of rewriting the expected value in terms of $\sum_{x=1} ^ \infty f(x)$. Any help is appreciated. Thank you.

After looking at the multiple ways to solving this, I am going with the summation direction. This is what I have done so far, but I am still not there yet. Any correction of the following is appreciated.

New:

$\sum_{x=1}^\infty \frac {x}{2^x}= \sum_{x=0}^\infty \frac{x+1}{2^{x+1}}=\sum_{x=0}^\infty \frac {x}{2^{x+1}}+ \sum_{x=0}^\infty \frac {1}{2^{x+1}}= \frac {1}{2} \sum_{x=0}^ \infty \frac {x}{2^x}+ \frac {1}{2} \sum_{x=0}^ \infty \frac {1}{2^x}$

From here, I do not see how $\frac {1}{2} \sum_{x=0}^ \infty \frac {1}{2^x}=1$?