calculate-n-0-1-n-2-4-

Question Number 144500 by mathmax by abdo last updated on 25/Jun/21

calculate \sum_{n = 0}^{\infty} \frac{1}{n^{2} + 4}

Answered by Dwaipayan Shikari last updated on 26/Jun/21

\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} + a^{2}} = \frac{π}{2 a} c o t h (\frac{π}{a}) - \frac{1}{2 a^{2}}

\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} + 4} = \frac{π}{4} c o t h (\frac{π}{2}) - \frac{1}{8}

\underset{n = 0}{\sum^{\infty}} \frac{1}{n^{2} + 4} = \frac{π}{4} \coth (\frac{π}{2}) + \frac{7}{8}

Answered by ArielVyny last updated on 25/Jun/21

w e h a v e f (t) = \frac{1}{t^{2} + 4} p o s i t i v a n d d e c r e a s i n g

i n [0 . \infty] t h e n

\int_{0}^{\infty} \frac{1}{t^{2} + 4} d t = \int_{0}^{\infty} \frac{1}{4 (1 + \frac{t^{2}}{4})} d t = \frac{1}{4} \int_{0}^{\infty} \frac{1}{1 + \frac{t^{2}}{4}} d t

u = \frac{t}{4} d u = \frac{1}{4} d t

\int_{0}^{\infty} \frac{1}{1 + u^{2}} d u = {[a r c t g]}_{0}^{\infty} = \frac{π}{2}

Commented by mathmax by abdo last updated on 26/Jun/21

the Q is a sum not integral sir viny \dots