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Limit ratio test

  1. Jul 10, 2013 #1
    1. The problem statement, all variables and given/known data

    $$ \sum_{n=1} ^\infty\frac{1} {2^n} $$

    2. Relevant equations

    3. The attempt at a solution
    I know just by looking at it that it converges no problem. You do the ratio test and you get something of the form

    \displaystyle\lim_{n\rightarrow \infty} {\frac{2^n}{2^{n+1}}}

    If $$2^n$$ goes to infinity. And $$2^{n+1}$$ does also. Then how is this equal to 1/2 ?It should be infinity over infinity. Which I know it can't be since I already know it converges by looking.
  2. jcsd
  3. Jul 10, 2013 #2
    Before you take the limit, simplify [itex]\frac{2^{n}}{2^{n+1}}[/itex].
  4. Jul 10, 2013 #3


    Staff: Mentor

    $$\lim_{n\to \infty} \frac{2^n}{2^{n+1}}= \lim_{n\to \infty} \frac{2^n}{2 \cdot 2^n} = \frac 1 2$$
  5. Jul 10, 2013 #4
    GRRRRRR!!!!!!!!!!!!!! Thanks.
  6. Jul 10, 2013 #5


    User Avatar
    Science Advisor

    In fact, except for the fact that the sum starts at n= 1 rather than n= 0, that's a geometric series. It is easy to show that [tex]\sum_{n=0}^\infty r^n= \frac{1}{1- r}[/tex]. Calculate that with r=1/2 then subtract off the value when n= 0, which is [tex](1/2)^0= 1[/tex].
  7. Jul 10, 2013 #6
    Yeah I saw it was geometric they just wanted me to to use the ratio test. It is funny you mention this. I want to know why does it matter whether you start and n =1 or n =0. ?
  8. Jul 10, 2013 #7
    For checking convergence, it doesn't matter whether you start with 0 or 1.
    If you want to find the actual value of the sum, then it does matter because

    [tex]\sum_{n=0}^{+\infty} r^n = 1 + \sum_{n=1}^{+\infty} r^n[/tex]

    So the results are different.
  9. Jul 10, 2013 #8
    The original was ##\sum_{n=1} ^\infty\frac{1} {2^n}##

    Why would you not just do

    ##\sum_{n=1}^\infty r^n= \frac{1}{1- r}##

    notice n starts at 1. Why did Halls of Ivy change it then subtract off the one? I mean I get why but why not just start with the original at n = 1 and sum? Does my question make any sense?
  10. Jul 10, 2013 #9

    [tex]\sum_{n=1}^\infty r^n= \frac{1}{1- r}[/tex]

    is not true. The equality is only true when ##n## starts with ##0##.

    That said, we do have

    [tex]\sum_{n=1}^\infty r^n= \frac{1}{1- r} = -1+\sum_{n=0}^\infty r^n = -1+\frac{1}{1-r} = \frac{r}{1-r}[/tex]
    Last edited: Jul 10, 2013
  11. Jul 10, 2013 #10
    You wrote ..$$\sum_{n=1}^\infty r^n= \frac{1}{1- r} = -1+\sum_{n=1}^\infty r^n = -1+\frac{1}{1-r} = \frac{r}{1-r}$$

    Did you mean...

    $$\sum_{n=0}^\infty r^n= \frac{1}{1- r} = -1+\sum_{n=1}^\infty r^n = -1+\frac{1}{1-r} = \frac{r}{1-r}$$
    ? Or I'm mistaken?
    Thank you !
  12. Jul 10, 2013 #11
    Neither. I have corrected the post. Thanks for spotting the error!
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