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

Question Number 154441 by talminator2856791 last updated on 18/Sep/21

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

Commented by benhamimed last updated on 18/Sep/21

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

= l i m \frac{2^{2} \times 3^{2} \times 4^{2} \times . . \times {(n - 1)}^{2} \times n^{2} \times {(n + 1)}^{2}}{(1 \times 3) (2 \times 4) (3 \times 5) \times . . \times ((n - 2) n) ((n - 1) (n + 1)) (n (n + 2))}

l i m \frac{2 (n + 1)}{(n + 2)} = 2

Answered by Kamel last updated on 18/Sep/21

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

P = \underset{n = 1}{\prod^{+ \infty}} \frac{n + 1}{n} . \frac{n + 1}{n + 2} = \frac{2}{1} . \frac{2}{3} . \frac{3}{2} . \frac{3}{4} . \frac{4}{3} . \frac{4}{5} \dots

= 2

Commented by mnjuly1970 last updated on 18/Sep/21

l n (p) = l i m \underset{k = 1}{\sum^{n}} {l n (\frac{k + 1}{k}) + l n (\frac{k + 1}{k + 2})}

= {l i m}_{n \to \infty} (n + 1) + l n (2) - {l i m}_{n \to \infty} (n + 2)

= {l i m}_{n \to \infty} (\frac{n + 1}{n + 2}) + l n (2) = l n (2)

∴ p = 2