prove-S-n-1-1-1-3-2pi-3-27-

Question Number 152345 by mnjuly1970 last updated on 27/Aug/21

p r o v e \dots

S = \underset{n = 1}{\sum^{\infty}} \frac{1}{} = \frac{1}{3} + \frac{2 π \sqrt{3}}{27} \dots

Answered by Kamel last updated on 27/Aug/21

p r o v e \dots

S = \underset{n = 1}{\sum^{\infty}} \frac{1}{} = \frac{1}{3} + \frac{2 π \sqrt{3}}{27} \dots

= \underset{n = 1}{\sum^{+ \infty}} \frac{n Γ (n + 1) Γ (n)}{Γ (2 n + 1)} = \underset{n = 1}{\sum^{+ \infty}} n \int_{0}^{1} t^{n - 1} {(1 - t)}^{n} d t

= \int_{0}^{1} \frac{d}{d y} {[\underset{n = 1}{\sum^{+ \infty}} y^{n} {(1 - t)}^{n}]}_{y = t} d t = \int_{0}^{1} \frac{1 - t}{{(1 - t (1 - t))}^{2}} d t

= \int_{0}^{1} \frac{u}{{(u^{2} - u + 1)}^{2}} d u = \int_{1}^{+ \infty} \frac{u}{{(u^{2} - u + 1)}^{2}} d u

S = \frac{1}{2} \int_{0}^{+ \infty} \frac{u d u}{{(u^{2} - u + 1)}^{2}}

= \frac{1}{2} \frac{d}{d a} {[\int_{0}^{+ \infty} \frac{d u}{(u^{2} - a u + 1)}]}_{a = 1} = \frac{1}{2} \frac{d}{d a} {[\frac{1}{\sqrt{1 - \frac{a^{2}}{4}}} (\frac{π}{2} + A r c t a n (\frac{a}{2 \sqrt{1 - \frac{a^{2}}{4}}}))]}_{a = 1}

= \frac{1}{2} [\frac{2}{3 \sqrt{3}} (\frac{π}{2} + A r c t a n (\frac{1}{\sqrt{3}})) - \frac{1}{2 \sqrt{3}} (\frac{\frac{- 1}{\sqrt{3}} - \sqrt{3}}{4})]

= \frac{2 π}{9 \sqrt{3}} + \frac{1}{3} = \frac{1}{3} + \frac{2 π \sqrt{3}}{27}

Commented by mnjuly1970 last updated on 28/Aug/21

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