S-x-n-1-2n-2n-1-x-2n-x-lt-1-

Question Number 144924 by qaz last updated on 30/Jun/21

S (x) = \underset{n = 1}{\sum^{\infty}} \frac{(2 n)!!}{(2 n + 1)!!} x^{2 n} = ? \dots \dots . . (∣ x ∣< 1)

Answered by Ar Brandon last updated on 30/Jun/21

(2 n + 1)!! = \frac{(2 n + 1)!}{(2 n)!!}

S (x) = \underset{n = 1}{\sum^{\infty}} \frac{{((2 n)!!)}^{2}}{(2 n + 1)!} x^{2 n} = \underset{n = 1}{\sum^{\infty}} \frac{2^{2 n} {(n!)}^{2}}{(2 n + 1)!} x^{2 n}

= \underset{n = 1}{\sum^{\infty}} {(2 x)}^{2 n} β (n + 1, n + 1) = \int_{0}^{1} \underset{n = 1}{\sum^{\infty}} {(2 x)}^{2 n} t^{n} {(1 - t)}^{n} dt

= \int_{0}^{1} \underset{n = 1}{\sum^{\infty}} {({4 x}^{2} t - {4 x}^{2} t^{2})}^{n} dt = \int_{0}^{1} \frac{dt}{{4 x}^{2} t^{2} - {4 x}^{2} t + 1} - 1

= \frac{1}{{4 x}^{2}} \int_{0}^{1} \frac{dt}{{(t - \frac{1}{2})}^{2} + \frac{1 - x^{2}}{{4 x}^{2}}} - 1 = \frac{1}{{4 x}^{2}} \cdot \frac{2 ∣ x ∣}{\sqrt{1 - x^{2}}} {[\tan^{- 1} (\frac{2 ∣ x ∣ t - ∣ x ∣}{\sqrt{1 - x^{2}}})]}_{0}^{1} - 1

= \frac{∣ x ∣}{x^{2} \sqrt{1 - x^{2}}} \tan^{- 1} (\frac{∣ x ∣}{\sqrt{1 - x^{2}}}) - 1

Commented by qaz last updated on 30/Jun/21

i found {it}^{'} s complicate to use DE to solve it \dots .

thank you sir

Commented by Ar Brandon last updated on 30/Jun/21

Je vous en prie !