advanced-calculus-prove-that-n-0-1-n-4n-2-1-1-4-2-picsch-pi-2-

Question Number 137307 by mnjuly1970 last updated on 31/Mar/21

\dots . . a d v a n c e d c a l c u l u s \dots . .

p r o v e t h a t ::

\underset{n = 0}{\sum^{\infty}} (\frac{{(- 1)}^{n}}{4 n^{2} + 1}) = \frac{1}{4} (2 + π c s c h (\frac{π}{2}))

\dots \dots \dots \dots \dots \dots \dots \dots \dots .

Answered by Dwaipayan Shikari last updated on 31/Mar/21

1 - \frac{1}{5} + \frac{1}{17} - \frac{1}{37} + \frac{1}{65} - \frac{1}{101} + \dots

1, 17, 65

g_{n} = {a n}^{2} + b n + c \Rightarrow g_{0} = c = 1

g_{1} = a + b + c = 17 \Rightarrow a + b = 16

g_{2} = 4 a + 2 b + 1 = 65 \Rightarrow 2 a + b = 32 \Rightarrow a = 16 b = 0 c = 1 \Rightarrow g_{n} = 16 n^{2} + 1

5, 37, 101 g_{n} = {(4 n + 2)}^{2} + 1

\underset{n = 0}{\sum^{\infty}} \frac{1}{16 n^{2} + 1} - \frac{1}{16 n^{2} + 16 n + 5}

= \frac{1}{16} (\frac{π}{2 . \frac{1}{4}} c o t h (\frac{π}{4}) - \frac{1}{2 . \frac{1}{16}}) - \frac{1}{4 i \sqrt{5}} \underset{n = 0}{\sum^{\infty}} \frac{1}{(n - (\frac{- 4 + i \sqrt{5}}{8}))} - \frac{1}{(n - (\frac{- 4 - i \sqrt{5}}{8}))}

= \frac{π}{8} c o t h (\frac{π}{4}) - \frac{1}{2} - \frac{1}{4 i \sqrt{5}} (ψ (\frac{1}{2} + \frac{i \sqrt{5}}{4}) - ψ (\frac{1}{2} - \frac{i \sqrt{5}}{8}))

= \frac{π}{8} c o t h (\frac{π}{4}) - \frac{1}{2} + \frac{π}{4 i \sqrt{5}} c o t (\frac{1}{2} + \frac{i \sqrt{5}}{8}) π

= \frac{π}{8} c o t h (\frac{π}{4}) - \frac{1}{2} - \frac{π}{4 \sqrt{5}} t a n h (\frac{\sqrt{5}}{8} π)

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