Calculus

posted by andre

Calculate the interval of convergence.
Sigma from n=1 to infinity of (n!*x^n)/n^n.
Calculus - Steve, Friday, July 29, 2016 at 11:52pm
try using the ratio test. You should be able to show that the series converges for |x| < e

I tried and I got
|x|lim n-infinity of |(n/n+1)^n|
which I get |x|<1

  1. Steve

    Using the ratio test,

    An+1/An =

    (n+1)!/(n+1)^(n+1)
    --------------------- x
           n!/n^n

    = (n+1)!/n! n^n/(n+1)^(n+1) x
    = (n+1) (n/(n+1))^n * 1/(n+1) x
    = (n/(n+1))^n x
    Hmmm. what to do?
    = ((n+1)/n)^-n x
    = (1 + 1/n)^-n x
    This should look familiar
    (1 + 1/n)^-n --> 1/e
    So, now we have the ratio test saying that

    An+1/An = 1/e x

    For the series to converge, the ratio must be less than 1, which means that

    |x| < e

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