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

Question Number 80036 by jagoll last updated on 30/Jan/20

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

Commented by john santu last updated on 30/Jan/20

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

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

= \frac{1}{2} {\frac{1}{2.3} - \frac{1}{3.4} + \frac{1}{3.4} - \frac{1}{4.5} + \frac{1}{4.5} - \frac{1}{5.6} + \dots}

the last term equat to zero,

all in between terms are cancelled

∴ \underset{n = 1}{\sum^{\infty}} \frac{1}{(n + 1) (n + 2) (n + 3)} = \frac{1}{12}

Commented by mr W last updated on 30/Jan/20

t h e q u e s t i o n w a s o r i g i n a l l y

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

w h i c h h a s t h e r e s u l t \frac{1}{4} .

Commented by jagoll last updated on 31/Jan/20

w a w \dots i h a v e t w o s o l u t i o n

(1) \underset{n = 1}{\sum^{\infty}} \frac{n}{(n + 1) (n + 2) (n + 3)} = \frac{1}{4}

(2) \underset{n = 1}{\sum^{\infty}} \frac{1}{(n + 1) (n + 2) (n + 3)} = \frac{1}{12}

t h a n k s y o u

Answered by Kamel Kamel last updated on 30/Jan/20

\frac{n}{(n + 1) (n + 2) (n + 3)} = \frac{A}{n + 1} + \frac{B}{n + 2} + \frac{C}{n + 3} \dots (E)

(E) \times (n + 1), n = - 1 \Rightarrow A = - \frac{1}{2}

(E) \times (n + 2), n = - 2 \Rightarrow B = 2

(E) \times (n + 3), n = - 3 \Rightarrow C = - \frac{3}{2}

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

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

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

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