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

Question Number 45237 by maxmathsup by imad last updated on 10/Oct/18

c a l c u l a t e \sum_{n = 2}^{\infty} \frac{1}{n^{3} - n}

Commented by maxmathsup by imad last updated on 11/Oct/18

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

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

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

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

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

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

b u t \sum_{k = 2}^{n} \frac{1}{k} = H_{n} - 1

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

\sum_{k = 2}^{n} \frac{1}{k + 1} = \sum_{k = 3}^{n + 1} \frac{1}{k} = H_{n + 1} - \frac{3}{2} \Rightarrow S_{n} = - H_{n + 1} + 1 + \frac{1}{2} H_{n - 1} + \frac{1}{2} H_{n + 1} - \frac{3}{4}

\Rightarrow S_{n} = \frac{1}{2} (l n (n - 1) + γ + l n (n + 1) + γ + o (\frac{1}{n})) - l n (n) - γ + o (\frac{1}{n}) + \frac{1}{4}

= l n (\sqrt{n - 1} \sqrt{n + 1}) - l n (n) + o (\frac{1}{n}) + \frac{1}{4}

= l n {\frac{\sqrt{n^{2} - 1}}{n}} + o (\frac{1}{n}) + \frac{1}{4} \to \frac{1}{4} (n \to + \infty) s o {l i m}_{n \to + \infty} S_{n} = \frac{1}{4} .

Answered by Smail last updated on 11/Oct/18

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

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

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

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

= \frac{1}{2} (1 + \frac{1}{2} + \frac{1}{3} + \dots) + \frac{1}{2} (\frac{1}{3} + \frac{1}{4} + \dots) - (\frac{1}{2} + \frac{1}{3} + \dots)

= \frac{1}{2} (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{3} + \frac{1}{4} + \frac{1}{4} + \dots) - (\frac{1}{2} + \frac{1}{3} + \dots)

= \frac{1}{2} (1 + \frac{1}{2}) + (\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \dots) - \frac{1}{2} - (\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \dots)

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

Commented by Tawa1 last updated on 11/Oct/18

Sir please what is the sum of nth term of \frac{1}{1.2.3} + \frac{1}{4.5.6} + \frac{1}{7.8.9} + \dots = ?

Commented by maxmathsup by imad last updated on 11/Oct/18

y o u r a n s w e r i s c o r r e c t s i r s m a 3 l t b a n k s . .