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

Question Number 26877 by abdo imad last updated on 30/Dec/17

f i n d t h e v a l u e o f \sum_{n = 1}^{\propto} \frac{1}{n (n + 1) (n + 2) (n + 3)} .

Commented by abdo imad last updated on 31/Dec/17

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

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

a = {l i m}_{x - > 0} x F (x) = \frac{1}{6}, b = {l i m}_{x - > - 1} (x + 1) F (x) = - \frac{1}{2}

c = {l i m}_{x - > - 2} (x + 2) F (x) = \frac{1}{2}, d = {l i m}_{x - > - 3} (x + 3) F (x) = - \frac{1}{6}

\Rightarrow F (x) = \frac{1}{6 x} - \frac{1}{2 (x + 1)} + \frac{1}{2 (x + 2)} - \frac{1}{6 (x + 3)}

l e t p u t S_{n} = \sum_{k = 1}^{n} \frac{1}{k (k + 1) (k + 2) (k + 3)}

S_{n} = \frac{1}{6} \sum_{k = 1}^{n} \frac{1}{k} - \frac{1}{2} \sum_{k = 1}^{n} \frac{1}{k + 1} + \frac{1}{2} \sum_{k = 1}^{k = n} \frac{1}{k + 2} - \frac{1}{6} \sum_{k = 1}^{n} \frac{1}{k + 3}

b u t \sum_{k = 1}^{n} \frac{1}{k} = H_{n}

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

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

\sum_{k = 1}^{k = n} \frac{1}{k + 3} = \sum_{k = 4}^{n + 3} \frac{1}{k} = H_{n + 3} - \frac{11}{6}

S_{n} = \frac{1}{6} H_{n} - \frac{1}{2} (H_{n + 1} - 1) + \frac{1}{2} (H_{n + 2} - \frac{3}{2}) - \frac{1}{6} (H_{n + 3} - \frac{11}{6})

= \frac{1}{6} (H_{n} - H_{n + 3}) + \frac{1}{2} (H_{n + 2} - H_{n + 1}) + \frac{1}{2} - \frac{3}{4} + \frac{11}{36}

b u t {l i m}_{n - >\propto} (H_{n} - H_{n + 3}) = 0 a n d {l i m}_{n - >\propto} (H_{n + 2} - H_{n + 1}) = 0

\Rightarrow {l i m}_{n - >\propto} S_{n} = - \frac{1}{4} + \frac{11}{36} = \frac{- 9 + 11}{36} = \frac{2}{36} = \frac{1}{18} .

Commented by abdo imad last updated on 31/Dec/17

t h i s i s t h e m e t h o d o f H_{n}

Answered by prakash jain last updated on 31/Dec/17

\frac{1}{n (n + 1) (n + 2) (n + 3)}

= \frac{A}{n} + \frac{B}{n + 1} + \frac{C}{n + 2} + \frac{D}{n + 3}

1 = A (n + 1) (n + 2) (n + 3) + B n (n + 2) (n + 3)

+ C n (n + 1) (n + 3) + D n (n + 1) (n + 2)

n = 0 \Rightarrow A = \frac{1}{6}

n = - 1 \Rightarrow B = - \frac{1}{2}

n = - 2 \Rightarrow C = \frac{1}{2}

n = - 3 \Rightarrow D = - \frac{1}{6}

= \frac{1}{6 n} - \frac{1}{2 (n + 1)} + \frac{1}{2 (n + 2)} - \frac{1}{6 (n + 3)}

\frac{1}{6.1} - \frac{1}{2 \cdot 2} + \frac{1}{2 \cdot 3} - \frac{1}{6 \cdot 4}

\frac{1}{6.2} - \frac{1}{2 \cdot 3} + \frac{1}{2 \cdot 4} - \frac{1}{6 \cdot 5}

\frac{1}{6.3} - \frac{1}{2 \cdot 4} + \frac{1}{2 \cdot 5} - \frac{1}{6 \cdot 6}

\frac{1}{6.4} - \frac{1}{2 \cdot 5} + \frac{1}{2 \cdot 6} - \frac{1}{6 \cdot 7}

\frac{1}{6 (m - 2)} - \frac{2}{2 (m - 1)} + \frac{1}{2 (m)} - \frac{1}{6 (m + 1)}

\frac{1}{6 (m - 1)} - \frac{2}{2 (m)} + \frac{1}{2 (m + 1)} - \frac{1}{6 (m + 2)}

\frac{1}{6 (m)} - \frac{2}{2 (m + 1)} + \frac{1}{2 (m + 2)} - \frac{1}{6 (m + 3)}

sum to m term

\frac{1}{6.1} + \frac{1}{6.2} + \frac{1}{6.3} - \frac{1}{2.2}

+ \frac{1}{2 (m + 2)} - \frac{1}{6 (m + 1)} - \frac{1}{6 (m + 2)} - \frac{1}{6 (m + 3)}

= \frac{1}{18} + \frac{3 (m + 1) (m + 3) - (m + 2) (m + 3) - (m + 1) (m + 3) - (m + 1) (m + 2)}{6 (m + 1) (m + 2) (m + 3)}

= \frac{1}{18} + \frac{2 (m + 1) (m + 3) - (m + 2) (m + 3) - (m + 1) (m + 2)}{6 (m + 1) (m + 2) (m + 3)}

= \frac{1}{18} - \frac{1}{3 (m + 1) (m + 2) (m + 3)}

sum to infinity =

\frac{1}{6 \cdot 1} + \frac{1}{6.2} + \frac{1}{6.3} - \frac{1}{2.2} = \frac{6 + 3 + 2 - 9}{6 \cdot 2 \cdot 3} = \frac{1}{18}