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Math Help - Help in finding the sum of the series!

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    Post Help in finding the sum of the series!

    Find the sum of: -

    A) (-2/3)^k when K goes from 1 to infinity...I got (3/5)!

    B) 2^k / [3^(2k+1) when k goes from 0 to infinity...I got (9/7)!

    c) 1 / k(k+1)...when K goes from 1 to infinity...I got (1)

    d) 1 / k(k+2)...when K goes from 1 to infinity...I am not sure how to split this term in two different term.



    I have got the solution for starting three, but I am not sure if I have done it right or not. I used the geometric series formula (1/1-r) in order to solve them.
    Please help me making it sure!

    Thanks!
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    Quote Originally Posted by Vedicmaths View Post
    Find the sum of: -

    A) (-2/3)^k when K goes from 1 to infinity...I got (3/5)!

    B) 2^k / [3^(2k+1) when k goes from 0 to infinity...I got (9/7)!

    c) 1 / k(k+1)...when K goes from 1 to infinity...I got (1)

    d) 1 / k(k+2)...when K goes from 1 to infinity...I am not sure how to split this term in two different term.



    I have got the solution for starting three, but I am not sure if I have done it right or not. I used the geometric series formula (1/1-r) in order to solve them.
    Please help me making it sure!

    Thanks!
    For d) you can use partial fraction decomp with this form

    \frac{1}{k(k+2)}=\frac{A}{k}+\frac{B}{K+2}

    Or note that

    \frac{1}{k(k+2)}=\frac{1}{2}\left( \frac{2}{k(k+2)}\right)

    \frac{1}{2}\left( \frac{k+2-k}{k(k+2)}\right)=\frac{1}{2}\left( \frac{k+2}{k(k+2)}\right)+\frac{1}{2}\left( \frac{-k}{k(k+2)}\right)

    \frac{1}{2}\left( \frac{1}{k}\right)+\frac{1}{2}\left( \frac{-1}{(k+2)}\right)

    writing out a few terms of each of the sums gives

    \frac{1}{2} \left( 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...\right)
    -\frac{1}{2}\left(\frac{1}{3}+\frac{1}{4}+...\right  )

    by telescoping we get

    \frac{1}{2}\left( 1+\frac{1}{2}\right)=\frac{3}{4}
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  3. #3
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    Hi !

    B) 2^k / [3^(2k+1) when k goes from 0 to infinity...I got (9/7)!
    Not quite

    \frac{2^k}{3^{2k+1}}=\frac 13 \cdot \frac{2^k}{3^{2k}}=\frac 13 \cdot \frac{2^k}{9^k}

    Just to note that you forgot \frac 13
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