A-nice-series-If-n-2-1-n-2-n-1-pi-tan-api-b-find-the-value-of-b-a-

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} = ?

Commented by Hridiana last updated on 21/Oct/23

\frac{3}{a - 1} + \int 5 - b d b

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{⏟}{}}

Commented by Hridiana last updated on 21/Oct/23

4 + 1 x \sqrt{x + 1}

Commented by HomeAlone last updated on 21/Oct/23