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!

Homework Help: Simplify the series

  1. Jul 31, 2012 #1
    1. The problem statement, all variables and given/known data
    This is a sub problem, the whole problem is find interval of convergence and radius of convergence but the thing im having trouble with is how to simplify this
    simplify the series


    2. Relevant equations

    3. The attempt at a solution

    keys say 2*4*6*8...(2n)=2^n*n!

    so 2^n*n!/2^(n+1)*(n+1)!=1/(2n+2)

    my question is, HOW??? how do u go from 2*4*6*8...(2n) to 2^n*n! or how do u get from

    2*4*6*8*(2n)/2*4*6*8....(2n+2) to 1/(2n+2)

    sorry for not using the proper tools, they are very confusing :(
  2. jcsd
  3. Jul 31, 2012 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    I'll address, "HOW??? how do u go from 2*4*6*8...(2n) to 2^n*n! ?" .

    [itex]=(2\cdot1)(2\cdot2)(2\cdot3)(2\cdot4)(2\cdot5)(2 \cdot6)(2\cdot7)\dots2(n-1)\cdot2(n)[/itex]

    [itex]=2^n\cdot1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7 \dots(n-1)\cdot(n)[/itex]

  4. Jul 31, 2012 #3


    User Avatar
    Science Advisor
    Homework Helper

    For example, 2*4*6*8=(2*1)*(2*2)*(2*3)*(2*4)=(2*2*2*2)*(1*2*3*4)=2^4*4!. Just think about it a little. It's not magic.
  5. Jul 31, 2012 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    2*4*6*8...*(2n) = (1*2)*(2*2)*(3*2)*(4*2)...(n*2) The first factor in each parentheses gives the n! and the second gives 2n.
    In the denominator the factors increase by 2 each factor. What would the factor right in front of the (2n+2) be?

    [Edit]Guess I have to learn to type faster. While I'm typing, two other responses appear.
  6. Aug 2, 2012 #5
    thanks for replies! appreciate it
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook