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Question Number 119442 by mnjuly1970 last updated on 24/Oct/20

\dots a d v a n c e d c a l c u l u s \dots

p r o v e t h a t : \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} (\begin{matrix} 2 n \\ n \end{matrix})} \overset{? ? ?}{=} \frac{ζ (2)}{3}

s o l u t i o n :: \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} * \frac{(2 n)!}{{(n!)}^{2}}} = \underset{n = 1}{\sum^{\infty}} \frac{n! * n!}{n^{2} * (2 n)!}

= \underset{n = 1}{\sum^{\infty}} \frac{Γ (n) Γ (n + 1)}{n Γ (2 n + 1)} = \underset{n = 1}{\sum^{\infty}} \frac{β (n, n + 1)}{n}

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

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

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

= - \int_{0}^{1} \frac{l n (1 + x^{3}) - l n (1 + x)}{x} d x

= \int_{0}^{1} \frac{l n (1 + x)}{x} d x - \int_{0}^{1} \frac{l n (1 + x^{3})}{x} d x

= - {l i}_{2} (- 1) - \int_{0}^{1} \frac{\sum_{n = 1} \frac{{(- 1)}^{n + 1} x^{3 n}}{n}}{x} d x

= \frac{π^{2}}{12} + \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n} \int_{0}^{1} x^{3 n - 1} d x

= \frac{π^{2}}{12} + \frac{1}{3} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{2}} = \frac{π^{2}}{12} - \frac{π^{2}}{36}

= \frac{π^{2}}{18} = \frac{ζ (2)}{3} ✓ ✓

m . n . j u l y . 1970 . .