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Question Number 112217 by mathdave last updated on 06/Sep/20

p r o p o r s e d b y m . n j u l y 1970

e v a l u a t e

\underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{n 2^{n}}

s o l u t i o n

r e c a l l t h a t \underset{n = 1}{\sum^{\infty}} H_{n} x^{n} = \frac{\ln (1 - x)}{x - 1}

∵ \frac{H_{n} x^{n}}{n} = \int_{0}^{x} H_{n} t^{n - 1} d t

\underset{n = 1}{\sum^{\infty}} \int_{0}^{x} H_{n} t^{n - 1} d t = \int_{0}^{x} \frac{\ln (1 - t)}{t (t - 1)} d t

\underset{n = 1}{\sum^{\infty}} \int_{0}^{x} H_{n} t^{n - 1} d t = \int_{0}^{x} \frac{\ln (1 - t)}{t - 1} d t - \int_{0}^{x} \frac{\ln (1 - t)}{t} d t = A - B

l e t A = \int_{0}^{x} \frac{\ln (1 - t)}{t - 1} d t = - \int_{0}^{x} \frac{\ln (1 - t)}{1 - t} d t

A = - {[- \ln^{2} (1 - t)]}_{0}^{x} + \int_{0}^{x} - \frac{\ln (1 - t)}{1 - t} d t

2 A = \ln^{2} (1 - x)

A = \frac{1}{2} \ln^{2} (1 - x) \dots \dots (1)

t h e n

B = \int_{0}^{x} \frac{\ln (1 - t)}{t} d t = \int_{0}^{1} \frac{\ln (1 - t)}{t} d t + \int_{1}^{x} \frac{\ln (1 - t)}{t} d t

B = - {L i}_{2} (1) - [{L i}_{2} (x) - {L i}_{2} (1)]

B = - {L i}_{2} (x) \dots . (2)

b u t I = A - B

I = \underset{n = 1}{\sum^{\infty}} \int_{0}^{x} H_{n} t^{n - 1} d t = \frac{1}{2} \ln^{2} (2) + {L i}_{2} (x)

∵ \underset{n = 1}{\sum^{\infty}} \frac{H_{n} x^{n}}{n} = \frac{1}{2} \ln^{2} (2) + {L i}_{2} (x)

b u t x = \frac{1}{2}

\underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{n 2^{n}} = \frac{1}{2} \ln^{2} (2) + {L i}_{2} (\frac{1}{2}) = \frac{1}{2} \ln^{2} (2) + \frac{π^{2}}{12} - \frac{1}{2} \ln^{2} (2)

∵ \underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{n 2^{n}} = \frac{π^{2}}{12}

b y m a t h d a v e (06 / 09 / 2020)

Commented by mnjuly1970 last updated on 06/Sep/20

t h a n k y o u s o m u c h m r

m a t h d a v e \dots g r a t e f u l .

Commented by Tawa11 last updated on 06/Sep/21

great sir

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