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

Question Number 105571 by bobhans last updated on 30/Jul/20

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

Answered by bubugne last updated on 30/Jul/20

\frac{1}{(n + 1) (n + 2) (n + 3)} = \frac{0.5}{n + 1} - \frac{1}{n + 2} + \frac{0.5}{n + 3}

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

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

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

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

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

S_{n} = - \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{1}{n + 1} + 2 \underset{n = 0}{\sum^{\infty}} \frac{1}{n + 2} - \frac{3}{2} \underset{n = - 1}{\sum^{\infty}} \frac{1}{n + 3} - 1 + \frac{3}{2} \underset{n = - 1}{\sum^{0}} \frac{1}{n + 3}

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

S_{n} = - 1 + \frac{3}{4} + \frac{1}{2}

S_{n} = \frac{1}{4}

Commented by bobhans last updated on 30/Jul/20

j o o s s a n d c o o l l

Answered by Ar Brandon last updated on 30/Jul/20

\frac{n}{(n + 1) (n + 2) (n + 3)} = \frac{a}{n + 1} + \frac{b}{n + 2} + \frac{c}{n + 3}

n \to - 1 \Rightarrow 2 a = - 1 \Rightarrow a = - \frac{1}{2}, n \to - 2 \Rightarrow b = 2

n \to - 3 \Rightarrow 2 c = - 3 \Rightarrow c = - \frac{3}{2}

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

S_{n} = - \frac{1}{4} + \frac{2}{3} - \frac{3}{8} for n = 1

- \frac{1}{6} + \frac{1}{2} - \frac{3}{10} for n = 2

- \frac{1}{8} + \frac{2}{5} - \frac{1}{4} for n = 3

- \frac{1}{10} + \frac{1}{3} - \frac{3}{14} for n = 4

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

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

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

S_{n} = - \frac{1}{4} + \frac{2}{3} - \frac{1}{6} - \frac{3}{2 (n + 1)} + \frac{2}{n + 2} - \frac{3}{2 (n + 3)}

Double subscripts: use braces to clarify

Answered by nimnim last updated on 30/Jul/20

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

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

= [\frac{1}{n + 2} - \frac{1}{n + 3}] + \frac{1}{2} [\frac{1}{n + 1} - \frac{1}{n + 3}] - [\frac{1}{n + 1} - \frac{1}{n + 2}]

= \underset{n = 1}{\sum^{\infty}} [\frac{1}{n + 2} - \frac{1}{n + 3}] + \underset{n = 1}{\sum^{\infty}} \frac{1}{2} [\frac{1}{n + 1} - \frac{1}{n + 3}] - \underset{n = 1}{\sum^{\infty}} [\frac{1}{n + 1} - \frac{1}{n + 2}]

n o t e : a l l a r e n t h t e r m s o f t e l e s c o p i n g s e r i e s

= \underset{n \to \infty}{l i m} [\frac{1}{3} - \frac{1}{n + 3}] + \frac{1}{2} \underset{n \to \infty}{l i m} [\frac{1}{2} + \frac{1}{3} - \frac{1}{n - 2} - \frac{1}{n + 3}] - \underset{n \to \infty}{l i m} [\frac{1}{2} - \frac{1}{n + 2}]

= [\frac{1}{3} - 0] + \frac{1}{2} [\frac{1}{2} + \frac{1}{3} - 0 - 0] - [\frac{1}{2} - 0]

= \frac{1}{3} + \frac{5}{12} - \frac{1}{2} = \frac{4 + 5 - 6}{12} = \frac{3}{12} = \frac{1}{4}

Answered by Dwaipayan Shikari last updated on 30/Jul/20

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

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

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

\lim_{n \to \infty} \frac{3}{2} (\frac{1}{3} - \frac{1}{n + 3}) - \frac{1}{2} \lim_{n \to \infty} (\frac{1}{2} - \frac{1}{n + 2})

\frac{3}{2} (\frac{1}{3} - 0) - \frac{1}{2} (\frac{1}{2} - 0) = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}

Commented by Ar Brandon last updated on 30/Jul/20

wow ! Better than mine.��