Series

**Krizalid** · December 8th 2009, 06:53 AM

Given $H_n=\sum_{i=1}^n\frac1i,$ then compute $\sum_{n=1}^\infty\frac{H_n}{(n+1)2^n}.$

**PaulRS** · December 8th 2009, 08:40 AM

We have: $ \sum\limits_{n = 1}^\infty {H_n \cdot x^n } = \left( {\tfrac{1} {{1 - x}}} \right) \cdot \log \left( {\tfrac{1} {{1 - x}}} \right) $ when $|x|<1$ $(*)$

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

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

$(*)$ Because we have: $ \left( {\sum\limits_{n = 0}^\infty {x^n } } \right) \cdot \left( {\sum\limits_{n = 1}^\infty {\tfrac{{x^n }} {n}} } \right) = \sum\limits_{n = 1}^\infty {\left( {\sum\nolimits_{k = 1}^n {\tfrac{1} {k}} } \right) \cdot x^n } $

**Krizalid** · January 26th 2011, 06:33 PM

when i saw first this problem, i got beaten, but i tried later, and i think i consider my solution a bit brute.

$\begin{aligned} \sum\limits_{n=1}^{\infty }{\sum\limits_{i=1}^{n}{\frac{1}{i(n+1)2^{n}}}}&=\ sum\limits_{i=1}^{\infty }{\sum\limits_{n=0}^{\infty }{\frac{1}{i\cdot 2^{i}(n+i+1)2^{n}}}} \\ & =\int_{0}^{1}{\int_{0}^{1/2}{\left\{ \frac{1}{x}\left( \sum\limits_{i=1}^{\infty }{(xy)^{i}} \right)\left( \sum\limits_{n=0}^{\infty }{\bigg( \frac{y}{2} \bigg)^{n}} \right) \right\}\,dx}\,dy} \\ & =\int_{0}^{1}{\int_{0}^{1/2}{\frac{2y}{(1-xy)(2-y)}\,dx}\,dy}, \\ & =2\int_{0}^{1}{\int_{0}^{y/2}{\frac{du\,dy}{(1-u)(2-y)}}} \\ & =2\int_{0}^{1}{\frac{\ln 2-\ln (2-y)}{2-y}\,dy} \\ & =\ln ^{2}2. \end{aligned}$

**TheCoffeeMachine** · June 29th 2011, 04:59 PM

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.

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