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

Question

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^^

Answer 1

$$\sum_{k=2}^{\infty} \frac{1}{2^k} = \sum_{k=0}^{\infty} \frac{1}{2^k} - (1 +\frac{1}{2}) = \frac{1}{1 - \frac{1}{2}} -\frac{3}{2} = \frac{1}{2}$$So the answer is $5\times\frac{1}{2} = \frac{5}{2}$

Answer 2

Actually,$$\sum_{k=2}^\infty2^{-k}=2^{-2}\sum_{k=0}^\infty2^{-k}=2^{-2}\times2=\frac12.$$

Answer 3

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: $$\sum^\infty_{k=2} = \frac{L}{4} = \frac{1}{2}$$