find-the-value-of-n-1-1-4n-4n-1-4n-2-4n-3-

Question Number 140405 by mnjuly1970 last updated on 07/May/21

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

Θ := \underset{n = 1}{\sum^{\infty}} \frac{1}{4 n . (4 n + 1) . (4 n + 2) . (4 n + 3)} = ?

Answered by qaz last updated on 07/May/21

f (n) = \frac{1}{4 n (4 n + 1) (4 n + 2) (4 n + 3)} = \frac{A}{4 n} + \frac{B}{4 n + 1} + \frac{C}{4 n + 2} + \frac{D}{4 n + 3}

A = \underset{4 n \to 0}{\lim 4} n f (n) = \frac{1}{6} B = \lim_{4 n \to - 1} (4 n + 1) f (n) = - \frac{1}{2}

C = \lim_{4 n \to - 2} (4 n + 2) f (n) = \frac{1}{2} D = \lim_{4 n \to - 3} (4 n + 3) f (n) = - \frac{1}{6}

Θ = \underset{n = 1}{\sum^{\infty}} [\frac{1}{6 \times 4 n} - \frac{1}{2 (4 n + 1)} + \frac{1}{2 (4 n + 2)} - \frac{1}{6 (4 n + 3)}]

= \underset{n = 1}{\sum^{\infty}} [\frac{1}{24 n} - \frac{1}{8 (n + \frac{1}{4})} + \frac{1}{8 (n + \frac{1}{2})} - \frac{1}{24 (n + \frac{3}{4})}]

= \frac{1}{24} ψ (\frac{7}{4}) + \frac{1}{8} ψ (\frac{5}{4}) - \frac{1}{8} ψ (\frac{3}{2})

= - \frac{1}{24} γ + \frac{11}{36} - \frac{π}{24} - \frac{1}{4} l n 2

Facebook Tweet Pin