nice-math-calculate-n-1-1-

Question Number 140789 by mnjuly1970 last updated on 12/May/21

\dots \dots . n i c e \dots . . m a t h \dots .

c a l c u l a t e :: Θ := \underset{n = 1}{\sum^{\infty}} \frac{1}{} = ? ?

Answered by Dwaipayan Shikari last updated on 12/May/21

\underset{n = 1}{\sum^{\infty}} \frac{1}{(\begin{matrix} 2 n \\ n \end{matrix})} = \underset{n = 1}{\sum^{\infty}} \frac{n Γ (n) Γ (n + 1)}{Γ (2 n + 1)} = \int_{0}^{1} Σ {n x}^{n} {(1 - x)}^{n} \frac{1}{1 - x} d x

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

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

= 0 - \frac{1}{2.3} \int_{0}^{1} \frac{1}{{(x - \frac{1}{2} - \frac{\sqrt{3} i}{2})}^{2}} + \frac{1}{{(x - \frac{1}{2} + \frac{\sqrt{3}}{2} i)}^{2}} - \frac{2}{(1 - x + x^{2})} d x

= \frac{1}{3} \int_{0}^{1} \frac{1}{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} + \frac{1}{2.3} (\frac{1}{\frac{1}{2} - \frac{\sqrt{3}}{2} i} + \frac{1}{\frac{1}{2} + \frac{\sqrt{3}}{2} i} + \frac{2}{\sqrt{3} i + 1} - \frac{2}{\sqrt{3} i - 1})

= \frac{2}{3 \sqrt{3}} (\frac{π}{6} + \frac{π}{6}) + \frac{1}{6} (1 + 1)

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

Commented by mnjuly1970 last updated on 12/May/21

t h a n k s a l o t \dots

Answered by Ar Brandon last updated on 12/May/21

S_{n} = \underset{n = 1}{\sum^{\infty}} \frac{1}{\overset{2 n}{} C_{n}} = \underset{n = 1}{\sum^{\infty}} \frac{n! \times n!}{(2 n)!} = \underset{n = 1}{\sum^{\infty}} \frac{Γ^{2} (n + 1)}{Γ (2 n + 1)} = \underset{n = 1}{\sum^{\infty}} n β (n + 1, n)

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

f (t) = \underset{n = 1}{\sum^{\infty}} t^{n} = \frac{t}{1 - t} \Rightarrow f^{'} (t) = \underset{n = 1}{\sum^{\infty}} {nt}^{n - 1} = \frac{1}{{(t - 1)}^{2}}

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

Ostrogradsky gives

S_{n} = \frac{1}{3} {[\frac{x + 1}{x^{2} - x + 1} + \int \frac{dx}{x^{2} - x + 1}]}_{0}^{1}

= {[\frac{1}{3} \cdot \frac{x + 1}{x^{2} - x + 1} + \frac{2}{3 \sqrt{3}} \arctan (\frac{2 x - 1}{\sqrt{3}})]}_{0}^{1}

= \frac{2}{3} - \frac{1}{3} + \frac{2}{3 \sqrt{3}} (\frac{π}{6} + \frac{π}{6}) = \frac{1}{3} + \frac{2 π}{9 \sqrt{3}}

Commented by mnjuly1970 last updated on 12/May/21

g r a t e f u l \dots