calculate-n-1-4n-1-n-2-3n-1-2-

Question Number 32998 by abdo imad last updated on 09/Apr/18

c a l c u l a t e \sum_{n = 1}^{\infty} \frac{4 n + 1}{n^{2} {(3 n + 1)}^{2}} .

Commented by abdo imad last updated on 15/Apr/18

l e t d e c o m p o s e F (x) = \frac{4 x + 1}{x^{2} {(3 x + 1)}^{2}}

F (x) = \frac{a}{x} + \frac{b}{x^{2}} + \frac{c}{3 x + 1} + \frac{d}{{(3 x + 1)}^{2}}

b = {l i m}_{x \to 0} x^{2} F (x) = 1

d = {l i m}_{x \to - \frac{1}{3}} {(3 x + 1)}^{2} F (x) = \frac{- \frac{4}{3} + 1}{\frac{1}{9}} = 9 . (- \frac{1}{3}) = - 3

F (x) = \frac{a}{x} + \frac{1}{x^{2}} + \frac{c}{3 x + 1} - \frac{3}{{(3 x + 1)}^{2}}

{l i m}_{x \to \infty} x F (x) = 0 = a + c \Rightarrow c = - a \Rightarrow

F (x) = \frac{a}{x} + \frac{1}{x^{2}} - \frac{a}{3 x + 1} - \frac{3}{{(3 x + 1)}^{2}}

F (1) = \frac{5}{16} = a + 1 - \frac{a}{4} - \frac{3}{16} = \frac{3}{4} a + \frac{13}{16} \Rightarrow

5 = 12 a + 13 \Rightarrow 12 a = 5 - 13 = - 8 \Rightarrow a = - \frac{8}{12} = - \frac{2}{3}

F (x) = \frac{- 2}{3 x} + \frac{1}{x^{2}} + \frac{2}{3 (3 x + 1)} - \frac{3}{{(3 x + 1)}^{2}}

\sum_{k = 1}^{n} \frac{4 k + 1}{k^{2} {(3 k + 1)}^{2}} = - \frac{2}{3} \sum_{k = 1}^{n} \frac{1}{k} + \sum_{k = 1}^{n} \frac{1}{k^{2}}

+ \frac{2}{3} \sum_{k = 1}^{n} \frac{1}{3 k + 1} - 3 \sum_{k = 1}^{n} \frac{1}{{(3 k + 1)}^{2}}

= \frac{2}{3} \sum_{k = 1}^{n} (\frac{1}{3 k + 1} - \frac{1}{k}) + \sum_{k = 1}^{n} \frac{1}{k^{2}} - 3 \sum_{k = 1}^{n} \frac{1}{{(3 k + 1)}^{2}}

b u t \sum_{k = 1}^{n} \frac{1}{3 k + 1} = \frac{1}{4} + \frac{1}{7} + \frac{1}{10} + \dots . + \frac{1}{3 n + 1}

\sum_{k = 1}^{n} \frac{1}{k} = \sum_{k = 3 p} (\dots) + \sum_{k = 3 p + 1} (\dots) + \sum_{k = 3 p + 2} (\dots)

= \sum_{p = 1}^{[\frac{n}{3}]} \frac{1}{3 p} + \sum_{k = 0}^{[\frac{n - 1}{3}]} \frac{1}{3 p + 1} + \sum_{p = 0}^{[\frac{n - 2}{3}]} \frac{1}{3 p + 2} \dots b e c o n t i n u e d \dots