find-the-value-of-n-1-1-n-1-H-2n-n-where-H-n-1-1-2-1-3-1-n-

Question Number 197819 by mnjuly1970 last updated on 30/Sep/23

f i n d t h e v a l u e o f :

ϕ = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1} H_{2 n}}{n} = ?

w h e r e, H_{n} = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}

Answered by witcher3 last updated on 04/Oct/23

\int_{0}^{1} - x^{n - 1} \ln (1 - x) dx = \frac{H_{n}}{n}

Σ {(- 1)}^{n - 1} \frac{H_{2 n}}{n} = 2 \sum_{n ⩾ 1} \int_{0}^{1} {(- 1)}^{n} x^{2 n - 2} \ln (1 - x)

= 2 \int_{0}^{1} - \frac{\ln (1 - x)}{(1 + x^{2})}

= - 2 \int_{0}^{\frac{π}{4}} \ln (\cos (t) - \sin (t)) - \ln (\cos (t)) dt

= - 2 \int_{0}^{\frac{π}{4}} \ln (\sqrt{2} \cos (t + \frac{π}{4})) - \ln (\cos (t)) dt

= - 2 \ln (\sqrt{2}) . \frac{π}{4} - 2 \int_{0}^{\frac{π}{4}} \ln (tg (t)) dt

= - \frac{\ln (2) π}{4} + 2 G

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