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

Question Number 113789 by Khalmohmmad last updated on 15/Sep/20

\underset{n = 1}{\sum^{\propto}} \frac{3}{n (n + 3)} = ?

Answered by aleks041103 last updated on 15/Sep/20

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

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

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

= (\underset{n = 1}{\sum^{\infty}} \frac{1}{n}) - (\underset{n = 4}{\sum^{\infty}} \frac{1}{n}) =

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

= \underset{n = 1}{\sum^{3}} \frac{1}{n} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = \frac{11}{6}

\Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{3}{n (n + 3)} = \frac{11}{6}

Answered by Dwaipayan Shikari last updated on 15/Sep/20

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

= (\frac{1}{3} - 0) + 2 \sum^{\infty} \frac{n + 1}{n (n + 1) (n + 2) (n + 3)}

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

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

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

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

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

= \frac{1}{2} + \frac{1}{9} = \frac{11}{18}

S o

\underset{n = 1}{\sum^{\infty}} \frac{1}{n (n + 3)} = \frac{11}{18} s o, \underset{n = 1}{\sum^{\infty}} \frac{3}{n (n + 3)} = \frac{11}{6}