prove-that-pi-cotan-pi-lim-n-k-n-n-1-k-

Question Number 67532 by mathmax by abdo last updated on 28/Aug/19

p r o v e t h a t π c o t a n (π α) = {l i m}_{n \to + \infty} \sum_{k = - n}^{n} \frac{1}{α - k}

Commented by ~ À ® @ 237 ~ last updated on 29/Aug/19

U s e t h e f o r m u l a s \underset{- \infty}{\sum^{\infty}} f (k) = - \sum_{z_{k}} R e s (f (z) π c o t a n (π z), z_{k}) a n d \underset{- \infty}{\sum^{\infty}} {(- 1)}^{k} f (k) = - \sum_{z_{k}} R e s (f (z) π c s c (π z), z_{k}) [w i t h c s c (π x) = \frac{1}{s i n (π x)}

S o h e r e f (k) = \frac{1}{α - k} a n d z_{k} = α

\underset{- \infty}{\sum^{\infty}} \frac{1}{α - k} = - R e s (\frac{π c o t a n (π z)}{α - z}, α) = - \lim_{z \to α} (\frac{(z - α) π c o t a n (π z)}{α - z}) = π c o t a n (π α)