Question-143031

Question Number 143031 by rs4089 last updated on 09/Jun/21

Answered by mnjuly1970 last updated on 09/Jun/21

\frac{π^{2}}{6} + {l n}^{2} (2) \dots .

Answered by Dwaipayan Shikari last updated on 09/Jun/21

S = \underset{n = 1}{\sum^{\infty}} H_{n}^{2} x^{n} = H_{1}^{2} x + H_{2}^{2} x^{2} + H_{3} x^{3} + \dots

S (1 - x) = x + (H_{2}^{2} - H_{1}^{2}) x^{2} + (H_{3}^{2} - H_{2}^{2}) x^{3} + \dots

S (1 - x) = x + \frac{H_{2} + H_{1}}{2} x^{2} + \frac{H_{2} + H_{3}}{3} x^{3} + \dots H_{n - 1} = H_{n} - \frac{1}{n}

S (1 - x) = \underset{n = 1}{\sum^{\infty}} \frac{H_{n} + H_{n - 1}}{n} x^{n} = \underset{n = 1}{\sum^{\infty}} \frac{2 H_{n} - \frac{1}{n}}{n} x^{n}

\Rightarrow S (1 - x) = 2 \underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{n} x^{n} - \underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{n^{2}}

\underset{n = 1}{\sum^{\infty}} H_{n} x^{n - 1} = - \frac{l o g (1 - x)}{x (1 - x)}

\underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{n 2^{n}} = - \int_{0}^{1 / 2} \frac{l o g (1 - x)}{x} d x - \int_{0}^{1 / 2} \frac{l o g (1 - x)}{1 - x} d x

= {L i}_{2} (\frac{1}{2}) + \frac{1}{2} {l o g}^{2} (2)

{L i}_{2} (1 - x) + {L i}_{2} (x) = \frac{π^{2}}{6} - \frac{1}{2} l o g (x) l o g (1 - x)

\Rightarrow {L i}_{2} (\frac{1}{2}) = \frac{π^{2}}{12} - \frac{1}{2} {l o g}^{2} (2)

S (1 - \frac{1}{2}) = 2 ({L i}_{2} (\frac{1}{2}) + {l o g}^{2} (2)) - {L i}_{2} (\frac{1}{2})

S = 2 {L i}_{2} (\frac{1}{2}) + 2 {l o g}^{2} (2) = \frac{π^{2}}{6} + {l o g}^{2} (2)

Commented by mnjuly1970 last updated on 09/Jun/21

g r e a t e \dots .