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

Question Number 33591 by abdo imad last updated on 19/Apr/18

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

Commented by abdo imad last updated on 20/Apr/18

l e t p u t S_{n} = \sum_{k = 1}^{n} \frac{2}{k^{3} + 3 k^{2} + 2 k} w e h a v e

k^{3} + 3 k^{2} + 2 k = k (k^{2} + 3 k + 2) = k (k^{2} + k + 2 k + 2)

= k (k (k + 1) + 2 (k + 1)) = k (k + 1) (k + 2) l e t d e c o m p o s e

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

a = {l i m}_{x \to 0} x F (x) = 1

b = {l i m}_{x \to - 1} (x + 1) F (x) = \frac{2}{(- 1) . 1} = - 2

c = {l i m}_{x \to - 2} (x + 2) F (x) = \frac{2}{(- 2) (- 1)} = 1 \Rightarrow

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

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

\sum_{k = 1}^{n} \frac{1}{k} = H_{n}

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

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

S_{n} = H_{n} + H_{n + 2} - 2 H_{n + 1} + 2 - \frac{3}{2} = H_{n} + H_{n + 2} - 2 H_{n + 1} + \frac{1}{2}

b u t H_{n} = l n (n) + γ + o (\frac{1}{n})

H_{n + 2} = l n (n + 2) + γ + o (\frac{1}{n})

H_{n + 1} = l n (n + 1) + γ + o (\frac{1}{n}) \Rightarrow

S_{n} = l n (n (n + 2)) + 2 γ - 2 l n (n + 1) - 2 γ + o (\frac{1}{n}) + \frac{1}{2}

= l n (\frac{n^{2} + 2 n}{{(n + 1)}^{2}}) + \frac{1}{2} \Rightarrow {l i m}_{n \to + \infty} S_{n} = \frac{1}{2} \Rightarrow

★ \sum_{n = 1}^{\infty} \frac{2}{n^{3} + 3 n^{2} + 2 n} = \frac{1}{2} ★

Answered by sma3l2996 last updated on 19/Apr/18

\underset{n = 1}{\sum^{\infty}} \frac{2}{n^{3} + 3 n^{2} + 2 n} = \underset{n = 1}{\sum^{\infty}} \frac{2}{n (n + 1) (n + 2)}

\frac{1}{n (n + 1) (n + 2)} = \frac{a}{n} + \frac{b}{n + 1} + \frac{c}{n + 2}

a = \frac{1}{2}; b = - 1; c = \frac{1}{2}

\underset{n = 1}{\sum^{\infty}} \frac{2}{n (n + 1) (n + 2)} = \underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{2}{n + 1} + \frac{1}{n + 2})

= \underset{n = 1}{\sum^{\infty}} \frac{1}{n} - \underset{n = 1}{\sum^{\infty}} \frac{2}{n + 1} + \underset{n = 1}{\sum^{\infty}} \frac{1}{n + 2}

l e t l = n + 1 a n d k = n + 2

\underset{n = 1}{\sum^{\infty}} \frac{2}{n (n + 1) (n + 2)} = \underset{n = 1}{\sum^{\infty}} \frac{1}{n} - \underset{l = 2}{\sum^{\infty}} \frac{2}{l} + \underset{k = 3}{\sum^{\infty}} \frac{1}{k}

= \underset{n = 1}{\sum^{\infty}} \frac{1}{n} - (\underset{l = 1}{\sum^{\infty}} \frac{2}{l} - \frac{2}{1}) + (\underset{k = 1}{\sum^{\infty}} \frac{1}{k} - \frac{1}{1} - \frac{1}{2})

= \underset{n = 1}{\sum^{\infty}} \frac{1}{n} - \underset{l = 1}{\sum^{\infty}} \frac{2}{l} + \underset{k = 1}{\sum^{\infty}} \frac{1}{k} + 2 - 1 - \frac{1}{2}

= \underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{2}{n} + \frac{1}{n}) + \frac{1}{2} = \frac{1}{2}