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Math Help - Evaluate this sum.

  1. #1
    PQR
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    Evaluate this sum.

    Find the sum: \sum_{n = 1}^{\infty}\frac{1}{\left(3n-1\right)\sqrt{2}\left(4n-1\right)}
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by PQR View Post
    Find the sum: \sum_{n = 1}^{\infty}\frac{1}{\left(3n-1\right)\sqrt{2}\left(4n-1\right)}
    Try partial fraction decomposition. See if you can obtain a telescoping sum.
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  3. #3
    PQR
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    Quote Originally Posted by VonNemo19 View Post
    Try partial fraction decomposition. See if you can obtain a telescoping sum.
    I have: \frac{3}{2\sqrt{2}}-\frac{4}{3\sqrt{2}}+\frac{3}{5\sqrt{2}}-\frac{4}{7\sqrt{2}}+\frac{3}{8\sqrt{2}}-\frac{4}{11\sqrt{2}}+\frac{3}{11\sqrt{2}}-\frac{4}{15\sqrt{2}}+...

    Not sure what to do with it, though.
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  4. #4
    Flow Master
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    Quote Originally Posted by PQR View Post
    I have: \frac{3}{2\sqrt{2}}-\frac{4}{3\sqrt{2}}+\frac{3}{5\sqrt{2}}-\frac{4}{7\sqrt{2}}+\frac{3}{8\sqrt{2}}-\frac{4}{11\sqrt{2}}+\frac{3}{11\sqrt{2}}-\frac{4}{15\sqrt{2}}+...

    Not sure what to do with it, though.
    Actually the problem is tougher than suggested by the first glance. I don't have time now - I will try to reply with some suggestions later if no-one else has done so.
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  5. #5
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    \frac{1}{(3n+2)(4n+3)}=\frac{3(4n+3)-4(3n+2)}{(3n+2)(4n+3)}=\frac{3}{3n+2}-\frac{4}{4n+3}, now use \frac1{n+1}=\int_0^1x^n\,dx and the expansion for the geometric series. Solve the remaining integral.

    (By the way, the general term changes because the series starts from n=0.)
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