1-2-pi-2-1-2-pi-2-1-6-pi-2-1-6-pi-2-1-10-pi-2-1-10-pi-2-pi-2-16-sec-2-pi-2-4-Prove-or-disprove-

Question Number 130555 by Dwaipayan Shikari last updated on 26/Jan/21

\frac{1}{{(2 - π)}^{2}} + \frac{1}{{(2 + π)}^{2}} + \frac{1}{{(6 - π)}^{2}} + \frac{1}{{(6 + π)}^{2}} + \frac{1}{{(10 - π)}^{2}} + \frac{1}{{(10 + π)}^{2}} + . . = \frac{π^{2}}{16} {s e c}^{2} (\frac{π^{2}}{4})

P r o v e o r d i s p r o v e

Answered by mindispower last updated on 27/Jan/21

S = \sum_{k ⩾ 0} \frac{1}{{(4 k + 2 - π)}^{2}} + \frac{1}{{(4 k + 2 + π)}^{2}} = \frac{1}{16} \sum_{k ⩾ 0} \frac{1}{{(k + \frac{2 - π}{4})}^{2}} + \frac{1}{{(k + \frac{2 + π}{4})}^{2}}

c o t (x) = \frac{1}{x} + \sum_{n ⩾ 1} \frac{1}{(x - π n)} + \frac{1}{(x + π n)}

\Rightarrow t g (π x) = c o t (\frac{π}{2} - x) = \frac{2}{π - 2 x} + \sum_{n ⩾ 1} \frac{2}{π - 2 x - 2 π n} + \frac{2}{π - 2 x + 2 π n}

\Rightarrow 1 + {t g}^{2} (x) = \frac{4}{{(π - 2 x)}^{2}} + \sum_{n ⩾ 1} \frac{4}{{(π - 2 x - 2 π n)}^{2}} + \frac{4}{{(π - 2 x + 2 π n)}^{2}} = H

x = \frac{π^{2}}{4}

\Rightarrow H = \frac{4}{{(π - \frac{π^{2}}{2})}^{2}} + \sum_{n ⩾ 1} \frac{4}{(π - \frac{π^{2}}{2} - 2 π n)} + \frac{4}{{(π - \frac{π^{2}}{2} + 2 π n)}^{2}} = A + B

B = \sum_{n ⩾ 0} {\frac{4}{{(- π - \frac{π^{2}}{2} - 2 n π)}^{2}} + \frac{4}{{(π + 2 π n - \frac{π^{2}}{2})}^{2}}} - \frac{4}{{(π - \frac{π^{2}}{2})}^{2}}

= \frac{1}{π^{2}} \sum_{n ⩾ 0} \frac{1}{{(\frac{2 + π}{4} + n)}^{2}} + \frac{1}{{(n + \frac{2 - π}{4})}^{2}} - \frac{4}{(π - \frac{π^{2}}{2})}

= \frac{16}{π^{2}} S - \frac{4}{{(π - \frac{π^{2}}{2})}^{2}} = {s e c}^{2} (\frac{π^{2}}{4}) - \frac{4}{{(π - \frac{π^{2}}{2})}^{2}}

\Rightarrow S = \frac{π^{2}}{16} {s e c}^{2} (\frac{π^{2}}{4})

Commented by Dwaipayan Shikari last updated on 27/Jan/21

G r e a t s i r! I h a d p o s t e d t h i s p r o b l e m o n B r i l l i a n t . Y o u c o u l d

c h e c k t h e i r a l s o

Commented by mindispower last updated on 27/Jan/21

w i t h e p l e a s u r s i r i h a v e n o t y e t B r i l l i a n t