Σ_(n=0) ^∞ (((2n)),(n) )(1/(4^n (2n+1)^3 ))=(π^3 /(48))+(π/4)ln^2 2


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Question Number 136733 by Ñï= last updated on 25/Mar/21

$\underset{n = 0}{\sum^{\infty}} (\begin{matrix} 2 n \\ n \end{matrix}) \frac{1}{4^{n} {(2 n + 1)}^{3}} = \frac{π^{3}}{48} + \frac{π}{4} {l n}^{2} 2$
	Answered by mindispower last updated on 25/Mar/21
	$start by (1/( (√(1−4x))))=Σ_(n≥0) (((2n)),(n) )x^n ⇒(1/( (√(1−4x^2 ))))Σ (((2n)),(n) )x^(2n) ⇒(1/2)arcsin(2x)=Σ (((2n)),(n) ).(x^(2n+1) /(2n+1)) ⇒(1/2)∫_0 ^t ((arcsin(2x))/x)dx=Σ (((2n)),(n) ).(t^(2n+1) /((2n+1)^2 )) we want ∫_0 ^(1/2) (1/x)∫_0 ^x ((arcsin(2t))/t)dtdx=Σ (((2n)),(n) ).(1/(4^n (2n+1)^3 )) by part[ ln(x)∫_0 ^x ((arcsin(2t))/t)dt]_0 ^(1/2) −∫_0 ^(1/2) ((ln(x)arcsin(2x))/x)dx =ln((1/2))∫_0 ^(1/2) ((arcsin(2x))/x)dx−[(1/2)ln^2 (x)arcsin(2x)]_0 ^(1/2) +∫_0 ^(1/2) ((ln^2 (x))/( (√(1−4x^2 ))))dx ∫_0 ^(1/2) ((arcsin(2x))/x)dx=∫_0 ^1 ((arcsin(t))/t)dt=−∫_0 ^1 ((ln(t))/( (√(1−t^2 ))))dt =−∫_0 ^(π/2) ln(sin(t))dt=((πln(2))/2) ∫_0 ^(1/2) ((ln^2 (x))/( (√(1−4x^2 ))))dx=(1/2)∫_0 ^1 ((ln^2 (t)−2ln(2)ln(t)+ln^2 (2))/( (√(1−t^2 )))) =(1/2){∫_0 ^1 ((ln^2 (t))/( (√(1−t^2 ))))−ln(2).((πln(2))/2)+ln^2 (2)(π/2)} =(1/2)A ∫_0 ^1 ((ln^2 (t))/( (√(1−t^2 ))))dt=∫_0 ^(π/2) ln^2 (sin(t))dt=A β(a,b)=2∫_0 ^(π/2) sin^(2a−1) (t)cos^(2b−1) (t)dt A=(1/8)(∂^2 /∂a^2 )β((3/2),(1/2)) ∂_a β=β(a,b)(Ψ(a)−Ψ(a+b)} ∂_a ^2 β=β(a,b){(Ψ(a)−Ψ(a+b))^2 +Ψ′(a)−Ψ′(a+b)} A=(1/8)β((3/2),(1/2)){(Ψ((1/2))−Ψ(2))^2 +Ψ′((1/2))−Ψ′(2)} A=(1/8)(Γ((3/2))Γ((1/2))){(π^2 /2)−(π^2 /6)+1+(−2ln(2)−γ−1+γ)^2 } A=(1/8).(1/2).π{(π^2 /3)+1+(−2ln(2)−1)^2 } S=Σ_(n≥0) (((2n)),(n) ).(1/(4^n (2n+1)^2 ))=−((πln^2 (2))/2)−((πln^2 (2))/4) +(π/(32))(4ln^2 (2)+2+4ln(2)+(π^2 /3))...continued later$
	$s t a r t b y$ $\frac{1}{\sqrt{1 - 4 x}} = \sum_{n ⩾ 0} (\begin{matrix} 2 n \\ n \end{matrix}) x^{n}$ $\Rightarrow \frac{1}{\sqrt{1 - 4 x^{2}}} Σ (\begin{matrix} 2 n \\ n \end{matrix}) x^{2 n}$ $\Rightarrow \frac{1}{2} a r c s i n (2 x) = Σ (\begin{matrix} 2 n \\ n \end{matrix}) . \frac{x^{2 n + 1}}{2 n + 1}$ $\Rightarrow \frac{1}{2} \int_{0}^{t} \frac{a r c s i n (2 x)}{x} d x = Σ (\begin{matrix} 2 n \\ n \end{matrix}) . \frac{t^{2 n + 1}}{{(2 n + 1)}^{2}}$ $w e w a n t \int_{0}^{\frac{1}{2}} \frac{1}{x} \int_{0}^{x} \frac{a r c s i n (2 t)}{t} d t d x = Σ (\begin{matrix} 2 n \\ n \end{matrix}) . \frac{1}{4^{n} {(2 n + 1)}^{3}}$ $b y p a r t {[l n (x) \int_{0}^{x} \frac{a r c s i n (2 t)}{t} d t]}_{0}^{\frac{1}{2}} - \int_{0}^{\frac{1}{2}} \frac{l n (x) a r c s i n (2 x)}{x} d x$ $= l n (\frac{1}{2}) \int_{0}^{\frac{1}{2}} \frac{a r c s i n (2 x)}{x} d x - {[\frac{1}{2} {l n}^{2} (x) a r c s i n (2 x)]}_{0}^{\frac{1}{2}}$ $+ \int_{0}^{\frac{1}{2}} \frac{{l n}^{2} (x)}{\sqrt{1 - 4 x^{2}}} d x$ $\int_{0}^{\frac{1}{2}} \frac{a r c s i n (2 x)}{x} d x = \int_{0}^{1} \frac{a r c s i n (t)}{t} d t = - \int_{0}^{1} \frac{l n (t)}{\sqrt{1 - t^{2}}} d t$ $= - \int_{0}^{\frac{π}{2}} l n (s i n (t)) d t = \frac{π l n (2)}{2}$ $\int_{0}^{\frac{1}{2}} \frac{{l n}^{2} (x)}{\sqrt{1 - 4 x^{2}}} d x = \frac{1}{2} \int_{0}^{1} \frac{{l n}^{2} (t) - 2 l n (2) l n (t) + {l n}^{2} (2)}{\sqrt{1 - t^{2}}}$ $= \frac{1}{2} {\int_{0}^{1} \frac{{l n}^{2} (t)}{\sqrt{1 - t^{2}}} - l n (2) . \frac{π l n (2)}{2} + {l n}^{2} (2) \frac{π}{2}}$ $= \frac{1}{2} A$ $\int_{0}^{1} \frac{{l n}^{2} (t)}{\sqrt{1 - t^{2}}} d t = \int_{0}^{\frac{π}{2}} {l n}^{2} (s i n (t)) d t = A$ $β (a, b) = 2 \int_{0}^{\frac{π}{2}} {s i n}^{2 a - 1} (t) {c o s}^{2 b - 1} (t) d t$ $A = \frac{1}{8} \frac{\partial^{2}}{\partial a^{2}} β (\frac{3}{2}, \frac{1}{2})$ $\partial_{a} β = β (a, b) (Ψ (a) - Ψ (a + b)}$ $\partial_{a}^{2} β = β (a, b) {{(Ψ (a) - Ψ (a + b))}^{2} + Ψ^{'} (a) - Ψ^{'} (a + b)}$ $A = \frac{1}{8} β (\frac{3}{2}, \frac{1}{2}) {{(Ψ (\frac{1}{2}) - Ψ (2))}^{2} + Ψ^{'} (\frac{1}{2}) - Ψ^{'} (2)}$ $A = \frac{1}{8} (Γ (\frac{3}{2}) Γ (\frac{1}{2})) {\frac{π^{2}}{2} - \frac{π^{2}}{6} + 1 + {(- 2 l n (2) - γ - 1 + γ)}^{2}}$ $A = \frac{1}{8} . \frac{1}{2} . π {\frac{π^{2}}{3} + 1 + {(- 2 l n (2) - 1)}^{2}}$ $S = \sum_{n ⩾ 0} (\begin{matrix} 2 n \\ n \end{matrix}) . \frac{1}{4^{n} {(2 n + 1)}^{2}} = - \frac{π {l n}^{2} (2)}{2} - \frac{π {l n}^{2} (2)}{4}$ $+ \frac{π}{32} (4 {l n}^{2} (2) + 2 + 4 l n (2) + \frac{π^{2}}{3}) . . . c o n t i n u e d l a t e r$
	Commented by Ñï= last updated on 25/Mar/21

	$t h a n k y o u s i r!$
	Commented by mindispower last updated on 26/Mar/21

	$p l e a s u r$
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