proporsed by m.n july 1970 evaluate Σ_(n=1) ^∞ (H_n /(n2^n )) solution recall thatΣ_(n=1) ^∞ H_n x^n =((ln(1−x))/(x−1)) ∵((H_n x^n )/n)=∫_0 ^x H_n t^(n−1) dt Σ_(n=1) ^∞ ∫_0 ^x H_n t^(n−1) dt=∫_0 ^x ((ln(1−t))/(t(t−1)))dt Σ_(n=1) ^∞ ∫_0 ^x H_n t^(n−1) dt=∫_0 ^x ((ln(1−t))/(t−1))dt−∫_0 ^x ((ln(1−t))/t)dt=A−B let A=∫_0 ^x ((ln(1−t))/(t−1))dt=−∫_0 ^x ((ln(1−t))/(1−t))dt A=−[−ln^2 (1−t)]_0 ^x +∫_0 ^x −((ln(1−t))/(1−t))dt 2A=ln^2 (1−x) A=(1/2)ln^2 (1−x)......(1) then B=∫_0 ^x ((ln(1−t))/t)dt=∫_0 ^1 ((ln(1−t))/t)dt+∫_1 ^x ((ln(1−t))/t)dt B=−Li_2 (1)−[Li_2 (x)−Li_2 (1)] B=−Li_2 (x)....(2) but I=A−B I=Σ_(n=1) ^∞ ∫_0 ^x H_n t^(n−1) dt=(1/2)ln^2 (2)+Li_2 (x) ∵Σ_(n=1) ^∞ ((H_n x^n )/n)=(1/2)ln^2 (2)+Li_2 (x) but x=(1/2) Σ_(n=1) ^∞ (H_n /(n2^n ))=(1/2)ln^2 (2)+Li_2 ((1/2))=(1/2)ln^2 (2)+(π^2 /(12))−(1/2)ln^2 (2) ∵Σ_(n=1) ^∞ (H_n /(n2^n ))=(π^2 /(12)) by mathdave(06/09/2020)


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

$p r o p o r s e d b y m . n j u l y 1970$ $e v a l u a t e$ $\underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{n 2^{n}}$ $s o l u t i o n$ $r e c a l l t h a t \underset{n = 1}{\sum^{\infty}} H_{n} x^{n} = \frac{\ln (1 - x)}{x - 1}$ $∵ \frac{H_{n} x^{n}}{n} = \int_{0}^{x} H_{n} t^{n - 1} d t$ $\underset{n = 1}{\sum^{\infty}} \int_{0}^{x} H_{n} t^{n - 1} d t = \int_{0}^{x} \frac{\ln (1 - t)}{t (t - 1)} d t$ $\underset{n = 1}{\sum^{\infty}} \int_{0}^{x} H_{n} t^{n - 1} d t = \int_{0}^{x} \frac{\ln (1 - t)}{t - 1} d t - \int_{0}^{x} \frac{\ln (1 - t)}{t} d t = A - B$ $l e t A = \int_{0}^{x} \frac{\ln (1 - t)}{t - 1} d t = - \int_{0}^{x} \frac{\ln (1 - t)}{1 - t} d t$ $A = - {[- \ln^{2} (1 - t)]}_{0}^{x} + \int_{0}^{x} - \frac{\ln (1 - t)}{1 - t} d t$ $2 A = \ln^{2} (1 - x)$ $A = \frac{1}{2} \ln^{2} (1 - x) . . . . . . (1)$ $t h e n$ $B = \int_{0}^{x} \frac{\ln (1 - t)}{t} d t = \int_{0}^{1} \frac{\ln (1 - t)}{t} d t + \int_{1}^{x} \frac{\ln (1 - t)}{t} d t$ $B = - {L i}_{2} (1) - [{L i}_{2} (x) - {L i}_{2} (1)]$ $B = - {L i}_{2} (x) . . . . (2)$ $b u t I = A - B$ $I = \underset{n = 1}{\sum^{\infty}} \int_{0}^{x} H_{n} t^{n - 1} d t = \frac{1}{2} \ln^{2} (2) + {L i}_{2} (x)$ $∵ \underset{n = 1}{\sum^{\infty}} \frac{H_{n} x^{n}}{n} = \frac{1}{2} \ln^{2} (2) + {L i}_{2} (x)$ $b u t x = \frac{1}{2}$ $\underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{n 2^{n}} = \frac{1}{2} \ln^{2} (2) + {L i}_{2} (\frac{1}{2}) = \frac{1}{2} \ln^{2} (2) + \frac{π^{2}}{12} - \frac{1}{2} \ln^{2} (2)$ $∵ \underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{n 2^{n}} = \frac{π^{2}}{12}$ $b y m a t h d a v e (06 / 09 / 2020)$
Commented by mnjuly1970 last updated on 06/Sep/20

$t h a n k y o u s o m u c h m r$ $m a t h d a v e . . . g r a t e f u l .$
Commented by Tawa11 last updated on 06/Sep/21

$great sir$
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