Calculate-k-1-H-k-2-k-k-1-where-H-k-is-the-k-th-harmonic-number-

Question Number 162478 by HongKing last updated on 29/Dec/21

Calculate : \underset{k = 1}{\sum^{\infty}} \frac{H_{k} 2^{- k}}{k + 1}

where H_{k} is the k - th harmonic number

Answered by mnjuly1970 last updated on 29/Dec/21

- - s o l u t i o n - -

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

i n t e g r a t i o n b o t h s i d e s

Σ H_{n} \int_{0}^{x} t^{n} d t = \frac{1}{2} {l n}^{2} (1 - x)

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

x = \frac{1}{2} \Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{(n + 1) . 2^{n}} = {l n}^{2} (\frac{1}{2})

Ψ = {l n}^{2} (2)

Commented by HongKing last updated on 31/Dec/21

Cool my dear Sir thank you so much