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Homework Help: Finding the general term of the series

  1. Sep 21, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the general term and test the nature(convergent/divergent) of:
    1/3 + 2/15 + 2/35 + .....

    3. The attempt at a solution
    If I simplify I get ,
    1/3 + (1)(2)/(3)(5) + (1)(2)(3)/(3)(5)(7) +.............

    I found that,
    TERM(N) = TERM(N-1)((N)/(2N+1)

    After this I am struck.
  2. jcsd
  3. Sep 21, 2014 #2
    The new format of this website is difficult for me.
  4. Sep 21, 2014 #3


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    Science Advisor

    You see, I suspect, that the numerator of the nth term is n! The denominator is something like a factorial except that it involves only odd integers- okay, make it a factorial by inserting the even integers:
    [tex]\frac{1}{3*5*7}= \frac{2*4*6}{2*3*4*5*6*7}= \frac{2*4*6}{7!}[/tex]

    Now, factor a 2 out of each term in the numerator: 2*4*6= 2(1)*2(2)*2*3= 2^3(3!).

    So [tex]\frac{1}{3*5*7}= \frac{2*4*6}{2*3*4*5*6*7}= \frac{2^3(3!)}{7!}[/tex]

    Your term [tex]\frac{1*2*3}{3*5*7}= \frac{2^3(3!)^2}{7!}[/tex].

    Can you write the nth term now?

    Do you see n
  5. Sep 21, 2014 #4
    yeah... i can write the nth term and i found that the series is convergent.

    But this series:
    1/4 + (1)(5)/(4)(8) + (1)(5)(9)/(4)(8)(12) + ...

    The general term for the denominator is 4^n (n!)

    but the numerator terms are tricky : 1*5*9

    ==> 1*2*3*4*5*6*7*8*9/2*3*4*6*7*8


    how do i simplify this part : 2*3*4*6*7*8 ?
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