Results 1 to 3 of 3

Math Help - Converging series with factorials

  1. #1
    Newbie
    Joined
    Dec 2008
    Posts
    2

    Converging series with factorials

    It's been a while since calculus and this problem actually is part of probability theory question. I've got the answer down into a series.

    ∑(1.5^n)/[n*(n-2!)] = ∑[(1.5^n)*(n-1)]/(n!)
    both summations from n=2 to infinity. (I proved that both sides are equal, just didn't know which would be easier to solve)

    I can see that the denominator will increase much faster than the numerator, leading me to believe that this series will converge. I also verified it by the ratio test and I also computed it in Excel. Without having Excel or a calculator, how can you solve this series by hand?

    From Excel, the series converges to 3.24084453516903.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by DannyG View Post
    It's been a while since calculus and this problem actually is part of probability theory question. I've got the answer down into a series.

    ∑(1.5^n)/[n*(n-2!)] = ∑[(1.5^n)*(n-1)]/(n!)
    both summations from n=2 to infinity. (I proved that both sides are equal, just didn't know which would be easier to solve)

    I can see that the denominator will increase much faster than the numerator, leading me to believe that this series will converge. I also verified it by the ratio test and I also computed it in Excel. Without having Excel or a calculator, how can you solve this series by hand?

    From Excel, the series converges to 3.24084453516903.
    Let us generalize \sum_{n=2}^{\infty}\frac{x^n}{n(n-2)!}=\sum_{n=2}^{\infty}\frac{(n-1)x^n}{n!}=\sum_{n=2}^{\infty}\frac{nx^n}{n!}-\sum_{n=2}^{\infty}\frac{x^n}{n!}=\sum_{n=0}^{\inf  ty}\frac{nx^n}{n!}-x-\left\{\sum_{n=0}^{\infty}\frac{x^n}{n!}-1-x\right\}

    Now consider that e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!} and consequently x\cdot e^x=\sum_{n=0}^{\infty}\frac{nx^n}{n!}

    So your sum is equal to

    \sum_{n=2}^{\infty}\frac{x^n}{n(n-2)!}=xe^x-x-\left(e^x-1-x\right)=xe^x-e^x+1

    So \sum_{n=2}^{\infty}\frac{\left(\frac{3}{2}\right)^  n}{n(n-2)!}=\frac{3}{2}e^{\frac{3}{2}}-e^{\frac{3}{2}}+1\approx 3.24408
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2008
    Posts
    2
    Thank you so much, that answers it perfectly.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. series (converging)
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: January 19th 2011, 03:44 AM
  2. Converging series
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 3rd 2010, 10:58 AM
  3. Converging series help
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 3rd 2010, 04:10 AM
  4. Series Problem With Factorials
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 1st 2010, 06:32 PM
  5. Converging series
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 12th 2008, 07:08 AM

Search Tags


/mathhelpforum @mathhelpforum