Infinite series- converges or not

  1. I have 2 questions I am having problems with.
    The goal is to determine if the series converges or not.

    Q1: Sum from(1 to inf) of (exp^i)/( (exp^2i) + 9)

    I tried to do the integral test but I cannot seem to integrate. Any guides would be appreciated.

    Also if I wanted to compare it what would I compare it to.

    Q2: Sum from (2 to inf) of 1/( (i^2) - 1)

    I did the integral test and came out to the following

    integral of 1/i^2 -1 = 1/2 (ln(i-1) - ln(i+1))

    So in looking at this as i approaches infinity, can I say that integral approaches 0 and therefore this series converges.

    All help is appreciated.


  2. jcsd
  3. Dick

    Dick 25,914
    Science Advisor
    Homework Helper

    There's an easy comparison for both of them. Q1) compare to exp(i)/exp(2i), Q2) compare to 2/(i^2).
  4. Tom Mattson

    Tom Mattson 5,526
    Staff Emeritus
    Science Advisor
    Gold Member

    The integral test is indeed a good way to go. Can you show us how you tried to integrate it? Once you do that, I'll point you in the right direction.

    The series does converge, but you can do better than the integral test. If you recognize that the series is telescoping, you can actually find its sum.
  5. Dick: thanks for response. I will look at your way. Always good to learn new ways

    Tom: This is how I did the integral test

    Let u = exp^(2x) + 9
    du/dx = 2exp^(2x)
    This is where I got stuck. I thought once I do du/dx, I should get e^x in this equation but since I didn't I couldn't get anywhere.

    So I tried by integration by parts and went into an infinite loop

    a- let u = 1/( exp(2x) + 9)
    b- du/dx = -2/(exp(2x +9)
    c- let dv = exp^x dx
    d- v = e^x

    To integrate
    uv - integral (v wrt dx)
    integral (e^x/e^2x +9) = e^x/(e^2x + 9) +2 integral (e^x/e^2x +9)

    This is where I got stuck.


  6. Tom Mattson

    Tom Mattson 5,526
    Staff Emeritus
    Science Advisor
    Gold Member

    You want to let u=exp(x) instead. You will end up with a rational function in u whose antiderivative you should recognize.
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