Easy series divergence/convergence problem

  1. Simfish

    Simfish 825
    Gold Member

    1. The problem statement, all variables and given/known data


    Test for convergence/divergence

    3. The attempt at a solution

    Using the Taylor expansion of e^x, we have...

    [tex]1+ \frac{1}{x} + \frac{1}{2!*x^2} + \frac{1}{3!*x^3} + ...[/tex]

    So as n -> infinity, we see that the function tends to [tex]1+ \frac{1}{x}[/tex]. Now we subtract 1 from this. So apparently the sum seems to act as according to the harmonic series, so it should be divergent. Is my reasoning correct? I tried solving it on maple, which refused to give a solution (perhaps since it was divergent).
  2. jcsd
  3. Gib Z

    Gib Z 3,347
    Homework Helper

    Using your nice Taylor Series expansion, just minus the 1 off that and we have:

    [tex]\sum_{x=1}^{\infty} \left( \frac{1}{x} + \frac{1}{2!x^2}+ {\frac{1}{3!x^3}..\right)[/tex].

    Now just use the integral test, which states that [tex]\sum_{x=1}^{\infty} f(x)[/tex] and [tex]\int^{\infty}_1 f(x)[/tex] converge together or diverge together if f(x) is continuous in the interval, positive and decreasing.

    So the integral is evaluated as such:
    [tex] \log_e x - \frac{1}{1\cdot 2!\cdot x} - \frac{1}{2\cdot 3! \cdot x^2} - \frac{1}{ 3\cdot 4! \cdot x^3} ...\bigg|_{1}^{\infty}[/tex].

    We can see that as x goes to infinity, all terms other than the log go to zero, whilst the log diverges.

    Hence the integral diverges, and so does your series.

    I'm not too sure about your previous reasoning, it seems a little disputable.
    Last edited: Jul 29, 2007
  4. Simfish

    Simfish 825
    Gold Member

    Ah yes, interesting test. My reasoning lies along the lines of...

    [tex]\sum_{x=1}^{\infty} \left( \frac{1}{x} + \frac{1}{2!x^2}+ {\frac{1}{3!x^3}..\right) > \sum_{x=1}^{\infty} \left( \frac{1}{x}\right)[/tex]. So since the harmonic series is divergent, and our series here is greater than the harmonic series for all positive x, our series is divergent by the limit comparison test.

    Is the integral test usually preferable to other tests if you can get the terms integrated? (since clearly you'd have to assume more in the comparison tests).
  5. Gib Z

    Gib Z 3,347
    Homework Helper

    Well actually no, maybe it was something about the way I read it in your first post, but re written in your last one, your chain of reasoning is perfectly fine :)

    I only use the integral test when I know that I can integrate the terms and if f(x) is continuous in the interval, positive and decreasing. It comes in handy more often than you would think actually. But it doesn't matter, usually theres a few ways to do these kinda questions.
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