# Finding sum of infinite series: sums of two series together

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1. Dec 22, 2016

### Ryaners

1. The problem statement, all variables and given/known data
Find the sum of the following series:
$$\left( \frac 1 2 + \frac 1 4 \right) + \left( \frac 1 {2^2} + \frac 1 {4^2} \right) +~...~+ \left( \frac 1 {2^k} + \frac 1 {4^k} \right) +~...$$

2. Relevant 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 $\sum_{n = 1}^{\infty} ar^k = \frac a {1-r}$

3. 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...!

2. Dec 22, 2016

### ehild

The formula is valid if the sum goes from k=0

3. Dec 22, 2016

### SammyS

Staff Emeritus
A couple of problems here.

First, the index in the Sigma notation should match the index in the items you are summing.

Then, as @ehild has pointed out, your index should start at zero, not 1 (unless you adjust for that).

Sum of a geometric series:
$\displaystyle \sum_{k = 0}^{\infty} ar^k = \frac a {1-r}$
or​
$\displaystyle \sum_{k = 1}^{\infty} ar^k = \frac {ar} {1-r}$

4. Dec 22, 2016

### Buffu

$$\sum_{n = 1}^{\infty} \frac 1 {2^k} = \sum_{n = 1}^{\infty} \left( \frac 1 2 \right)^k = \frac {\color{green}{1}} {1- \frac 1 2} = 2$$
The first term(a) is not 1.

5. Dec 22, 2016

### Ray Vickson

$$\sum_{n = 1}^{\infty} \frac 1 {2^k} = 2 \; \Longleftarrow \; \text{FALSE!}$$
$$\begin{array}{lcc} \frac{1}{2}+\frac{1}{4} &=& \frac{3}{4}\\ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} &=& \frac{7}{8}\\ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} +\frac{1}{16} &=& \frac{15}{16}\\ \vdots&\vdots&\vdots \end{array}$$
The sums are creeping up to 1, not to 2.

6. Dec 22, 2016

### Ryaners

Ah ok, my mistake. Thank you all for pointing that out, I didn't realise the formula would change if the index changed, though of course that makes sense. Cheers!

7. Dec 22, 2016

### SammyS

Staff Emeritus
I wouldn't consider it solved until you give a complete solution.

Yes, it's likely that now you do know how to solve this.​

8. Dec 22, 2016

### Buffu

Yes I know, that is the error in the post.
He forgot to account for first term of the GP (a) in the formula.

I should have wrote the formula in quote box.

9. Dec 23, 2016

### Ryaners

So by multiplying the formula by $r$ in order to start the index at 1 instead of 0, it becomes $\sum_{n=1}^{\infty} \frac {ar} {1-r}$ where $a = a_0 = 1$ for both series (because when summing a geometric series from $n=1$, the first term is $a_1 = ar$). The sum of each series is therefore 1 and $\frac 1 3$ respectively giving a total of $\frac 4 3$.

* she :)

Thanks again folks!

10. Dec 23, 2016

### Staff: Mentor

@Ryaners, please post questions about infinite series in the Calc & Beyond section. I have moved this thread.