Finding sum of infinite series: sums of two series together

[Ryaners]

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!

 

[ehild]

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 formula is valid if the sum goes from k=0

 

[SammyS]

...

Homework Equations

[/B]$$ \sum_{n = 1}^{\infty} \left( u_k+v_k \right) = \sum_{n = 1}^{\infty} u_k + \sum_{n = 1}^{\infty} v_k $$ $$ \sum_{n = 1}^{\infty} cu_k = c \sum_{n = 1}^{\infty} u_k $$ Sum of a geometric series $\sum_{n = 1}^{\infty} ar^k = \frac a {1-r}$

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​

$\displaystyle \sum_{k = 1}^{\infty} ar^k = \frac {ar} {1-r}$

 

[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.

 

[Ray Vickson]

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

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

 

[Ryaners]

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

 

[SammyS]

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.​

 

[Buffu]

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

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.

 

[Ryaners]

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.​

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 $\frac 4 3$.

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

* she :)

Thanks again folks!

 

[Mark44]

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