1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    [itex]
    \displaystyle\lim_{n\rightarrow \infty} {\frac{2^n}{2^{n+1}}}
    [/itex]


    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

    Mark44

    Staff: Mentor

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

    HallsofIvy

    User Avatar
    Staff Emeritus
    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Because

    [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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Neither. I have corrected the post. Thanks for spotting the error!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Limit ratio test
  1. Ratio test. (Replies: 2)

  2. Ratio Test (Replies: 1)

  3. Ratio Test (Replies: 3)

  4. Ratio Test (Replies: 5)

Loading...