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Question Number 103633 by Lordose last updated on 16/Jul/20

The question is

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

Commented by bobhans last updated on 16/Jul/20

w e l l w e l l

Answered by Dwaipayan Shikari last updated on 16/Jul/20

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

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

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

S_{1} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} {\sum^{\infty} \frac{1}{n} - \frac{1}{n + 1} = \underset{n \to \infty}{\lim 1} - \frac{1}{n + 1} = 1

S = \frac{3}{2} - \frac{3}{4} = \frac{3}{4} {\sum^{\infty} \frac{1}{n + 1} - \frac{1}{n + 2} = \lim_{n \to \infty} \frac{1}{2} - \frac{1}{n + 2} = \frac{1}{2}

Commented by mathmax by abdo last updated on 16/Jul/20

not correct

Commented by Dwaipayan Shikari last updated on 16/Jul/20

kindly can you show my mistake ?

Answered by Worm_Tail last updated on 16/Jul/20

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

p f d

A = Σ (\frac{3}{n + 1} - \frac{5}{2 (n + 2)} - \frac{1}{2 n})

A = 3 Σ (\frac{1}{n + 1}) - \frac{5}{2} Σ (\frac{1}{n + 2}) - \frac{1}{2} Σ (\frac{1}{n})

A = 3 (\frac{1}{2} + \sum_{n = 2} (\frac{1}{n + 1})) - \frac{5}{2} \sum_{n = 1} (\frac{1}{n + 2}) - \frac{1}{2} (1 + \frac{1}{2} + \sum_{n = 3} (\frac{1}{n}))

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

A = 3 (\frac{1}{2} + s) - \frac{5}{2} s - \frac{1}{2} (1 + \frac{1}{2} + s)

A = \frac{3}{2} + 3 s - \frac{5}{2} s - \frac{1}{2} - \frac{1}{4} - \frac{1}{2} s

A = 3 s - \frac{5}{2} s - \frac{1}{2} s + \frac{3}{4}

A = \frac{3}{4}

Commented by mathmax by abdo last updated on 16/Jul/20

correct answer thanks