Find the sum of the following geometric series $\sum_{k=2}^{\infty} \frac{5}{2^k} = \frac{5}{2}$ 
Find the sum of the following geometric series
$$\sum_{k=2}^{\infty} \frac{5}{2^k} = \frac{5}{2}$$

Attempt:
First I test with the root criterion if its divergent or convergent... $\frac {1}{2} < 1$ so it's convergent...
Now I try to find the sum and I can't get to the solution $\frac{5}{2}$.
Let's try:
$$\sum_{k=2}^{\infty} \frac{5}{2^k}=5\sum_{k=2}^{\infty} \frac{1}{2^k}=5\sum_{k=2}^{\infty} 2^{-^k}=5\cdot2^{-2}\sum_{k=2}^{\infty}2^{-k}$$
I don't know how to work with the $2^{-k}$ in here, hope for your help^^

You can combine the answers from S.H.W and José Carlos Santos to figure out a solution if you don't know the sum $\sum^\infty_{k=0} 2^{-k}$.
$$
\begin{align}
\sum^\infty_{k=2} 2^{-k} &= \sum^\infty_{k=0}2^{-k} - 1 - \frac{1}{2} \tag{S.H.W} \\
&=2^{-2}\sum^\infty_{k=0}2^{-k} \tag{José Carlos Santos}
\end{align}
$$
If we write $L=\sum^\infty_{k=0}$ then $L-\frac{3}{2} = \frac{L}{4}$. This has as only solution $L=2$. Therefore:
$$\sum^\infty_{k=2} = \frac{L}{4} = \frac{1}{2}$$
