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Math Help - Series

  1. #1
    Math Engineering Student
    Krizalid's Avatar
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    Series

    Given H_n=\sum_{i=1}^n\frac1i, then compute \sum_{n=1}^\infty\frac{H_n}{(n+1)2^n}.
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  2. #2
    Super Member PaulRS's Avatar
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    We have: <br />
\sum\limits_{n = 1}^\infty  {H_n  \cdot x^n }  = \left( {\tfrac{1}<br />
{{1 - x}}} \right) \cdot \log \left( {\tfrac{1}<br />
{{1 - x}}} \right)<br />
 when |x|<1 (*)

    Now integrate: <br />
\sum\limits_{n = 1}^\infty  {\tfrac{{H_n  \cdot x^{n + 1} }}<br />
{{n + 1}}}  = \int_0^x {\left( {\tfrac{1}<br />
{{1 - z}}} \right) \cdot \log \left( {\tfrac{1}<br />
{{1 - z}}} \right)dz}  = \int_0^x {\left( {\tfrac{1}<br />
{2} \cdot \log ^2 \left( {\tfrac{1}<br />
{{1 - z}}} \right)} \right)^\prime  dz} <br />

    Hence: <br />
\sum\limits_{n = 1}^\infty  {\tfrac{{H_n  \cdot x^{n + 1} }}<br />
{{n + 1}}}  = \tfrac{1}<br />
{2} \cdot \log ^2 \left( {\tfrac{1}<br />
{{1 - x}}} \right)<br />
for |x|<1. it follows then, by substituting x=\tfrac{1}{2} that <br />
\sum\limits_{n = 1}^\infty  {\tfrac{{H_n }}<br />
{{\left( {n + 1} \right) \cdot 2^n }}}  = \log ^2 \left( 2 \right)<br />

    (*) Because we have: <br />
\left( {\sum\limits_{n = 0}^\infty  {x^n } } \right) \cdot \left( {\sum\limits_{n = 1}^\infty  {\tfrac{{x^n }}<br />
{n}} } \right) = \sum\limits_{n = 1}^\infty  {\left( {\sum\nolimits_{k = 1}^n {\tfrac{1}<br />
{k}} } \right) \cdot x^n } <br />
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    when i saw first this problem, i got beaten, but i tried later, and i think i consider my solution a bit brute.

      \begin{aligned} \sum\limits_{n=1}^{\infty }{\sum\limits_{i=1}^{n}{\frac{1}{i(n+1)2^{n}}}}&=\  sum\limits_{i=1}^{\infty }{\sum\limits_{n=0}^{\infty }{\frac{1}{i\cdot 2^{i}(n+i+1)2^{n}}}} \\ <br />
 & =\int_{0}^{1}{\int_{0}^{1/2}{\left\{ \frac{1}{x}\left( \sum\limits_{i=1}^{\infty }{(xy)^{i}} \right)\left( \sum\limits_{n=0}^{\infty }{\bigg( \frac{y}{2} \bigg)^{n}} \right) \right\}\,dx}\,dy} \\ <br />
 & =\int_{0}^{1}{\int_{0}^{1/2}{\frac{2y}{(1-xy)(2-y)}\,dx}\,dy}, \\ <br />
 & =2\int_{0}^{1}{\int_{0}^{y/2}{\frac{du\,dy}{(1-u)(2-y)}}} \\ <br />
 & =2\int_{0}^{1}{\frac{\ln 2-\ln (2-y)}{2-y}\,dy} \\ <br />
 & =\ln ^{2}2.<br />
\end{aligned}
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  4. #4
    Super Member
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    Re: Series

    I propose a solution using Cauchy multiplication. Consider:

     \begin{aligned} & \ln^2(1-x)  = \bigg(\sum_{k \ge 0}\frac{x^{k+1}}{k+1}\bigg)^2 =  \bigg(\sum_{k \ge 0}\frac{x^{k+1}}{k+1}\bigg) \bigg(\sum_{k \ge 0}\frac{x^{k+1}}{k+1}\bigg) \\& = \sum_{k \ge 0} \sum_{0 \le n \le k} \bigg( \frac{x^{n+1}}{n+1}\cdot \frac{x^{k-n+1}}{k-n+1}\bigg)  =  \sum_{k \ge 0} \sum_{0 \le n \le k}\frac{x^{k+2}}{(n+1)(k-n+1)}\end{aligned}

    Now consider the inner sum in the following way:

    \begin{aligned} & \sum_{0 \le n \le k}\frac{1}{(n+1)(k-n+1)}  = \sum_{0 \le n \le k}\frac{(n+1)+(k-n+1)}{(k+2)(n+1)(k-n+1)} \\& = \sum_{0 \le n \le k }\frac{1}{(k+2)(n+1)}+\sum_{0 \le n \le k }\frac{1}{(k+2)(k-n+1)}\\& = \sum_{0 \le n \le k }\frac{1}{(k+2)(n+1)}+\sum_{0 \le n \le k }\frac{1}{(k+2)(n+1)} \\& =  \sum_{0 \le n \le k }\frac{2}{(k+2)(n+1)}.\end{aligned}

    Where on the third step we shifted the index for the sum on the right.


    \therefore \ln^2(1-x) = \sum_{k \ge 0} \sum_{0 \le n \le k }\frac{2x^{k+2}}{(k+2)(n+1)}

    Your sum is a special case of the above and occurs when x = 1/2.
    Last edited by TheCoffeeMachine; June 29th 2011 at 06:13 PM.
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