find-the-values-of-n-2-1-2-n-n-1-

Question Number 28981 by abdo imad last updated on 02/Feb/18

f i n d t h e v a l u e s o f \prod_{n = 2}^{\infty} (1 - \frac{2}{n (n + 1)}) .

Commented by abdo imad last updated on 03/Feb/18

\prod_{n = 2}^{\infty} (1 - \frac{2}{n (n + 1)}) = {l i m}_{n \to + \infty} S_{n} w i t h

S_{n} = \prod_{k = 2}^{n} (1 - \frac{2}{k (k + 1)}) = \prod_{k = 2}^{n} (\frac{k^{2} + k - 2}{k (k + 1)}) b u t

k^{2} + k - 2 = k^{2} - k + 2 k - 2 = k (k - 1) + 2 (k - 1)

= (k - 1) (k + 2)

S_{n} = \prod_{k = 2}^{n} \frac{k - 1}{k} . \frac{k + 2}{k + 1} = \prod_{k = 1}^{n - 1} \frac{k}{k + 1} . \prod_{k = 3}^{n + 1} \frac{k + 1}{k}

= \frac{1}{2} . \frac{2}{3} \prod_{k = 3}^{n - 1} \frac{k}{k + 1} . \prod_{k = 3}^{n - 1} \frac{k + 1}{k} . \frac{n + 1}{n} . \frac{n + 2}{n + 1}

S_{n} = \frac{1}{3} . \frac{n + 2}{n} \Rightarrow {l i m}_{n \to + \infty} S_{n} = \frac{1}{3} s o

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

Answered by iv@0uja last updated on 03/Feb/18

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

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

\underset{n = 2}{\prod^{m}} \frac{(n - 1) (n + 2)}{n (n + 1)}

= \frac{1 \cdot 4}{2 \cdot 3} \times \frac{2 \cdot 5}{3 \cdot 4} \times \frac{3 \cdot 6}{4 \cdot 5} \times \dots \times \frac{(m - 1) (m + 2)}{m (m + 1)}

= \frac{1}{3} \times \frac{m + 2}{m}

\lim_{m \to \infty} \frac{m + 2}{m} = 1

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

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