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Finding sum of infinite series: sums of two series together

  1. Dec 22, 2016 #1
    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 :redface:

    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!
     
  2. jcsd
  3. Dec 22, 2016 #2

    ehild

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    The formula is valid if the sum goes from k=0
     
  4. Dec 22, 2016 #3

    SammyS

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    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} ##
     
  5. Dec 22, 2016 #4
    $$\sum_{n = 1}^{\infty} \frac 1 {2^k} = \sum_{n = 1}^{\infty} \left( \frac 1 2 \right)^k = \frac {\color{green}{1}} {1- \frac 1 2} = 2$$
    The first term(a) is not 1.
     
  6. Dec 22, 2016 #5

    Ray Vickson

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    $$\sum_{n = 1}^{\infty} \frac 1 {2^k} = 2 \; \Longleftarrow \; \text{FALSE!}$$
    $$ \begin{array}{lcc} \frac{1}{2}+\frac{1}{4} &=& \frac{3}{4}\\
    \frac{1}{2} + \frac{1}{4} + \frac{1}{8} &=& \frac{7}{8}\\
    \frac{1}{2} + \frac{1}{4} + \frac{1}{8} +\frac{1}{16} &=& \frac{15}{16}\\
    \vdots&\vdots&\vdots
    \end{array}
    $$
    The sums are creeping up to 1, not to 2.
     
  7. Dec 22, 2016 #6
    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!
     
  8. Dec 22, 2016 #7

    SammyS

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    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.​
     
  9. Dec 22, 2016 #8
    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.
     
  10. Dec 23, 2016 #9
    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!
     
  11. Dec 23, 2016 #10

    Mark44

    Staff: Mentor

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