calculate-n-0-n-2-2n-

Question Number 147863 by mathmax by abdo last updated on 24/Jul/21

calculate \sum_{n = 0}^{\infty} \frac{n!^{2}}{(2 n)!}

Answered by qaz last updated on 24/Jul/21

\underset{n = 0}{\sum^{\infty}} \frac{{(n!)}^{2}}{(2 n)!}

= \underset{n = 0}{\sum^{\infty}} (2 n + 2) \int_{0}^{1} {(x - x^{2})}^{n} dx

= \int_{0}^{1} (2 yD + 2) ∣_{y = x - x^{2}} \underset{n = 0}{\sum^{\infty}} y^{n} dx

= \int_{0}^{1} (2 yD + 2) ∣_{y = x - x^{2}} \frac{1}{1 - y} dx

= \int_{0}^{1} \frac{2}{{(1 - y)}^{2}} ∣_{y = x - x^{2}} dx

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

= \frac{\frac{4}{3} x - \frac{2}{3}}{1 - x + x^{2}} ∣_{0}^{1} + \int_{0}^{1} \frac{\frac{4}{3}}{1 - x + x^{2}} dx

= \frac{4}{3} + \frac{8 \sqrt{3} π}{27}

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