Σ_(n=1) ^∞ (−1)^n ∙((2n−1)/((2n)!))∙((π/2))^(2n) =Σ_(n=1) ^∞ ((2n−1)/((2n)!))∙(−((π/2))^2 )^n =(2xD−1)∣_(x=π/2) Σ_(n=1) ^∞ (((−x^2 )^n )/((2n)!)) =(2xD−1)∣_(x=π/2) [Σ_(n=0) ^∞ (((−x^2 )^n )/((2n)!))−1] =(2xD−1)∣_(x=π/2) (cos x−1) =(−2xsin x−cos x+1)∣


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Question Number 142629 by qaz last updated on 03/Jun/21

$\underset{n = 1}{\sum^{\infty}} {(- 1)}^{n} \cdot \frac{2 n - 1}{(2 n)!} \cdot {(\frac{π}{2})}^{2 n}$ $= \underset{n = 1}{\sum^{\infty}} \frac{2 n - 1}{(2 n)!} \cdot {(- {(\frac{π}{2})}^{2})}^{n}$ $= (2 xD - 1) ∣_{x = π / 2} \underset{n = 1}{\sum^{\infty}} \frac{{(- x^{2})}^{n}}{(2 n)!}$ $= (2 xD - 1) ∣_{x = π / 2} [\underset{n = 0}{\sum^{\infty}} \frac{{(- x^{2})}^{n}}{(2 n)!} - 1]$ $= (2 xD - 1) ∣_{x = π / 2} (\cos x - 1)$ $= (- 2 xsin x - \cos x + 1) ∣_{x = π / 2}$ $= 1 - π$ $where is wrong ?$
	Answered by Dwaipayan Shikari last updated on 03/Jun/21

	$\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n} (2 n)}{(2 n)!} {(\frac{π}{2})}^{2 n} - \frac{{(- 1)}^{n}}{(2 n)!} {(\frac{π}{2})}^{2 n}$ $= - (\frac{2}{2!} {(\frac{π}{2})}^{2} - \frac{4}{4!} {(\frac{π}{2})}^{4} + \frac{6}{6!} {(\frac{π}{2})}^{6} + . . .) + (\frac{1}{2!} {(\frac{π}{2})}^{2} - \frac{1}{4!} {(\frac{π}{2})}^{4} + . .)$ $= - \frac{π}{2} (\frac{π}{2} - \frac{1}{3!} {(\frac{π}{2})}^{3} + . .) - (1 - \frac{1}{2!} {(\frac{π}{2})}^{2} - . . .) + 1$ $= - \frac{π}{2} s i n (\frac{π}{2}) - c o s (\frac{π}{2}) + 1 = 1 - \frac{π}{2}$
	Commented by qaz last updated on 03/Jun/21
	$i have another solution. Σ_(n=1) ^∞ (((−1)^n (2n−1))/((2n)!))x^(2n) ∣_(x=(π/2)) =Σ_(n=0) ^∞ (((−1)^(n+1) (2n+1))/((2n+2)!))x^(2n+2) ∣_(x=(π/2)) =−x^2 Σ_(n=0) ^∞ (((−1)^n )/((2n+2)(2n)!))x^(2n) =−x^2 ∫_0 ^1 yΣ_(n=0) ^∞ (((−1)^n )/((2n)!))∙(xy)^(2n) dy =−x^2 ∫_0 ^1 ycos (xy)dy =−x^2 {[(1/x)ysin (xy)+(1/x^2 )cos (xy)]_0 ^1 }_(x=(π/2)) =−x^2 {(1/x)sin x+(1/x^2 )(cos x−1)}_(x=(π/2)) =1−(π/2) −−−−−−−−−−−−−−−−−−− but Σ_(n=1) ^∞ (2n−1)(((−1)^n )/((2n)!))x^(2n) =(xD−1)(cos x−1), i still dont understand,why not LFT=(2xD−1)(cos x−1)???$
	$i have another solution .$ $\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n} (2 n - 1)}{(2 n)!} x^{2 n} ∣_{x = \frac{π}{2}}$ $= \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n + 1} (2 n + 1)}{(2 n + 2)!} x^{2 n + 2} ∣_{x = \frac{π}{2}}$ $= - x^{2} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{(2 n + 2) (2 n)!} x^{2 n}$ $= - x^{2} \int_{0}^{1} y \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{(2 n)!} \cdot {(xy)}^{2 n} dy$ $= - x^{2} \int_{0}^{1} ycos (xy) dy$ $= - x^{2} {{[\frac{1}{x} ysin (xy) + \frac{1}{x^{2}} \cos (xy)]}_{0}^{1}}_{x = \frac{π}{2}}$ $= - x^{2} {\frac{1}{x} \sin x + \frac{1}{x^{2}} (\cos x - 1)}_{x = \frac{π}{2}}$ $= 1 - \frac{π}{2}$ $- - - - - - - - - - - - - - - - - - -$ $but \underset{n = 1}{\sum^{\infty}} (2 n - 1) \frac{{(- 1)}^{n}}{(2 n)!} x^{2 n} = (xD - 1) (\cos x - 1),$ $i still dont understand, why not LFT = (2 xD - 1) (\cos x - 1) ? ? ?$
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