Obliczyć granicę ciągu

[Alik]

Mam do obliczenia granicę takiego ciągu:
\(\displaystyle{ \(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{4})...(1-\frac{1}{n})}\)
oraz
\(\displaystyle{ \(1-\frac{1}{2^{2}})(1-\frac{1}{3^{2}})(1-\frac{1}{4^{2}})...(1-\frac{1}{n^{2}})}\)
Z góry dzięki za wszelkie podpowiedzi

[Piotr Rutkowski]

Zauważ, że:
\(\displaystyle{ (1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{4})...(1-\frac{1}{n})=\frac{1}{2}*\frac{2}{3}*...*\frac{n-1}{n}=\frac{1}{n}}\)

[bolo]

\(\displaystyle{ \left(1-\frac{1}{2^{2}}\right)\cdot\left(1-\frac{1}{3^{2}}\right)\cdot\left(1-\frac{1}{4^{2}}\right)\cdot\left(1-\frac{1}{5^{2}}\right)\cdot\ldots\cdot\left(1-\frac{1}{n^{2}}\right)=\\= \frac{2^{2}-1}{2^{2}}\cdot\frac{3^{2}-1}{3^{2}}\cdot\frac{4^{2}-1}{4^{2}}\cdot\frac{5^{2}-1}{5^{2}}\cdot\ldots\cdot\frac{n^{2}-1}{n^{2}}=\\= \frac{(2-1)(2+1)}{2\cdot 2}\cdot\frac{(3-1)(3+1)}{3\cdot 3}\cdot\frac{(4-1)(4+1)}{4\cdot 4}\cdot\frac{(5-1)(5+1)}{5\cdot 5}\cdot\ldots\cdot\frac{(n-1)(n+1)}{n\cdot n}=\\= \frac{n+1}{2n}\stackrel{\footnotesize{n\to\infty}}{\longrightarrow}\frac{1}{2}}\)

[Alik]

Dzięki!!!