A nice series If , Ω = Σ_(n=2) ^∞ (( 1)/(n^( 2) +n −1)) =(( π tan( aπ ))/( b)) ⇒ find the value of (b/a) = ?


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Question Number 198496 by mnjuly1970 last updated on 21/Oct/23

$A n i c e s e r i e s$ $I f, Ω = \underset{n = 2}{\sum^{\infty}} \frac{1}{n^{2} + n - 1} = \frac{π t a n (a π)}{b}$ $\Rightarrow f i n d t h e v a l u e o f \frac{b}{a} = ?$
	Answered by mr W last updated on 21/Oct/23

	$Ω = \underset{n = 2}{\sum^{\infty}} \frac{1}{n^{2} + n - 1} = \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} + n - 1} - 1$ $\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} + n - 1}$ $= \underset{n = 1}{\sum^{\infty}} \frac{1}{(n - p) (n - q)} w i t h p, q = \frac{- 1 \pm \sqrt{5}}{2} a n d p + q = - 1, p q = - 1$ $= \frac{1}{p - q} \underset{n = 1}{\sum^{\infty}} (\frac{1}{n - p} - \frac{1}{n - q})$ $= \frac{1}{p - q} [ψ (1 - q) - ψ (1 - p)]$ $= \frac{1}{p - q} [ψ (1 + 1 + p) - ψ (1 - p)]$ $= \frac{1}{p - q} [ψ (1 + p) + \frac{1}{1 + p} - ψ (1 - p)]$ $= \frac{1}{q - p} [ψ (p) + \frac{1}{p} + \frac{1}{1 + p} - ψ (p) - π \cot p π]$ $= \frac{1}{p - q} [\frac{1}{p} + \frac{1}{1 + p} - π \cot p π]$ $= \frac{1}{p - q} [\frac{1}{p} - \frac{1}{q} - π \cot p π]$ $= \frac{1}{p - q} [\frac{q - p}{p q} - π \cot p π]$ $= - \frac{1}{p q} - \frac{π}{p - q} \cot p π$ $= 1 - \frac{π}{\sqrt{5}} \cot \frac{(- 1 + \sqrt{5}) π}{2}$ $= 1 + \frac{π}{\sqrt{5}} \cot (\frac{π}{2} - \frac{\sqrt{5} π}{2})$ $= 1 + \frac{π}{\sqrt{5}} \tan \frac{\sqrt{5} π}{2}$ $Ω = 1 + \frac{π}{\sqrt{5}} \tan \frac{\sqrt{5} π}{2} - 1$ $Ω = \frac{π}{\sqrt{5}} \tan \frac{\sqrt{5} π}{2} = \frac{π \tan (a π)}{b}$ $\Rightarrow a = \frac{\sqrt{5}}{2}, b = \sqrt{5} \Rightarrow \frac{b}{a} = 2$
	Commented by mnjuly1970 last updated on 21/Oct/23

	$s o n i c e s o l u t i o n s i r W$ $B r a v o \underset{―}{\underset{⏟}{}}$
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