find-n-1-1-n-2-n-1-3-

Question Number 68592 by Abdo msup. last updated on 14/Sep/19

f i n d \sum_{n = 1}^{\infty} \frac{1}{n^{2} {(n + 1)}^{3}}

Commented by Abdo msup. last updated on 16/Sep/19

l e t S_{n} = \sum_{k = 1}^{n} \frac{1}{k^{2} {(k + 1)}^{3}} w e h a v e S = {l i m}_{n \to + \infty} S_{n}

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

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

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

e = {l i m}_{x \to - 1} {(x + 1)}^{3} F (x) = 1 \Rightarrow

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

{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}{x + 1} + \frac{d}{{(x + 1)}^{2}} + \frac{1}{{(x + 1)}^{3}}

F (1) = \frac{1}{8} = a + 1 - \frac{a}{2} + \frac{d}{4} + \frac{1}{8} \Rightarrow

\frac{1}{2} a + \frac{d}{4} = - 1 \Rightarrow 2 a + d = - 4

F (- 2) = - \frac{1}{4} = - \frac{a}{2} + \frac{1}{4} + a + d - 1

= \frac{1}{2} a + d - \frac{3}{4} \Rightarrow - 1 = 2 a + 4 d - 3 \Rightarrow

2 a + 4 d = 2 \Rightarrow a + 2 d = 1 b u t d = - 2 a - 4 \Rightarrow

a - 4 a - 8 = 1 \Rightarrow - 3 a = 9 \Rightarrow a = - 3

\Rightarrow d = - 2 a - 4 = 6 - 4 = 2

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

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

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

\sum_{k = 1}^{n} \frac{1}{k + 1} = \sum_{k = 2}^{n + 1} \frac{1}{k} = H_{n + 1} - 1

\sum_{k = 1}^{n} \frac{1}{k^{2}} = ξ_{n} (2)

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

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

S_{n} = - \frac{3}{5} H_{n} + ξ_{n} (2) - \frac{3}{5} (H_{n + 1} - 1)

- \frac{2}{5} (ξ_{n + 1} (2) - 1) + ξ_{n + 1} (3) - 1 \Rightarrow

S_{n} = - 3 H_{n} + ξ_{n} (2) + 3 (H_{n + 1} - 1)

+ 2 (ξ_{n + 1} (2) - 1) + ξ_{n + 1} (3) - 1

= 3 (H_{n + 1} - H_{n}) + ξ_{n} (2) + 2 ξ_{n + 1} (2) + ξ_{n + 1} (3) - 6

b u t {l i m}_{n \to + \infty} (H_{n + 1} - H_{n}) = 0 \Rightarrow

{l i m}_{n \to + \infty} S_{n} = 3 ξ (2) + ξ (3) - 6

= 3 . \frac{π^{2}}{6} + ξ (3) - 6 = \frac{π^{2}}{2} + ξ (3) - 6 w i t h

ξ (3) \sim 1, 2

Commented by Abdo msup. last updated on 16/Sep/19

ξ_{n} (x) = \sum_{k = 1}^{n} \frac{1}{k^{x}}