find-the-value-of-the-following-series-n-1-cos-npi-4-n-2-

Question Number 191528 by mnjuly1970 last updated on 25/Apr/23

f i n d t h e v a l u e o f t h e

f o l l o w i n g s e r i e s .

Ω = \underset{n = 1}{\sum^{\infty}} \frac{c o s (\frac{n π}{4})}{n^{2}} = ?

Answered by witcher3 last updated on 25/Apr/23

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

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

= \frac{1}{2} ({Li}_{2} (e^{i \frac{π}{4}}) + {Li}_{2} (e^{- \frac{i π}{4}}))

{Li}_{2} (z) + {Li}_{2} (\frac{1}{z}) = - ζ (2) - \frac{1}{2} \ln^{2} (- z)

Ω = \frac{1}{2} ({Li}_{2} (e^{- \frac{i π}{4}}) + {Li}_{2} (\frac{1}{e^{- \frac{i π}{4}}}))

= \frac{1}{2} (- \frac{π^{2}}{6} - \frac{1}{2} \ln^{2} (- e^{- \frac{i π}{4}})) = \frac{1}{2} (- \frac{π^{2}}{6} + \frac{9 π^{2}}{32})

= \frac{11 π^{2}}{192}

Commented by mnjuly1970 last updated on 25/Apr/23

t h a n k s a l o t \dots . s i r

Commented by witcher3 last updated on 25/Apr/23

withe Pleasur Sir