- #1

Ryaners

- 50

- 2

## Homework Statement

Find the sum of the following series:

$$ \left( \frac 1 2 + \frac 1 4 \right) + \left( \frac 1 {2^2} + \frac 1 {4^2} \right) +~...~+ \left( \frac 1 {2^k} + \frac 1 {4^k} \right) +~...$$

## Homework Equations

$$ \sum_{n = 1}^{\infty} \left( u_k+v_k \right) = \sum_{n = 1}^{\infty} u_k + \sum_{n = 1}^{\infty} v_k $$

$$ \sum_{n = 1}^{\infty} cu_k = c \sum_{n = 1}^{\infty} u_k $$

Sum of a geometric series ## \sum_{n = 1}^{\infty} ar^k = \frac a {1-r} ##

## The Attempt at a Solution

The question referred back to the two relations I've put in the 'relevant equations' section - I assumed I'm meant to use the first one. So I split the sum into two sums as follows:

$$ \sum_{n = 1}^{\infty} \left( \frac 1 {2^k} + \frac 1 {4^k} \right) = \sum_{n = 1}^{\infty} \frac 1 {2^k} + \sum_{n = 1}^{\infty} \frac 1 {4^k} $$

For the left term, I got:

$$\sum_{n = 1}^{\infty} \frac 1 {2^k} = \sum_{n = 1}^{\infty} \left( \frac 1 2 \right)^k = \frac 1 {1- \frac 1 2} = 2$$

Similarly for the right term:

$$\sum_{n = 1}^{\infty} \frac 1 {4^k} = \sum_{n = 1}^{\infty} \left( \frac 1 4 \right)^k = \frac 1 {1- \frac 1 4} = \frac 4 3 $$

Then adding these together gives me ## 3 \frac 1 3 ##, which is... the wrong answer

I can see that ## \frac 1 {4^k} = \frac 1 {2^{2k}} ## - has that got something to do with the solution?

Edit: I should say that the answer in the back of the book is ## \frac 4 3 ## which makes me think that there must be a way that the first series is 'incorporated into' the second, but I can't quite get my head around that...!

Thanks in advance!