MCQ
The value of the limit $\prod\limits_{n = 2}^\infty {\,\left( {1 - \frac{1}{{{n^2}}}} \right)} $ is
- A$1$
- B$\frac{1}{4}$
- C$\frac{1}{3}$
- ✓$\frac{1}{2}$
✓
Answer
Correct option: D.
$\frac{1}{2}$
d
$\prod\limits_{n = 2}^\infty {\,\left( {\frac{{{n^2} - 1}}{{{n^2}}}} \right)} $=$\prod\limits_{n = 2}^\infty {\,\left( {\frac{{n - 1}}{n}} \right)} \,\,\prod\limits_{n = 2}^\infty {\,\left( {\frac{{n + 1}}{n}} \right)} $
=$\frac{{1\,\cdot\,2\,\cdot\,3........(n - 1)}}{{2\,\cdot\,3\,\cdot\,4......(n - 1)\,\cdot\,n}}$ =$\frac{1}{n}\,\cdot\,\frac{{n + 1}}{2}$ =$\mathop {Lim}\limits_{n \to \infty } \,\,\frac{{1 + \frac{1}{n}}}{2}$ =$\frac{1}{2}$
$\prod\limits_{n = 2}^\infty {\,\left( {\frac{{{n^2} - 1}}{{{n^2}}}} \right)} $=$\prod\limits_{n = 2}^\infty {\,\left( {\frac{{n - 1}}{n}} \right)} \,\,\prod\limits_{n = 2}^\infty {\,\left( {\frac{{n + 1}}{n}} \right)} $
=$\frac{{1\,\cdot\,2\,\cdot\,3........(n - 1)}}{{2\,\cdot\,3\,\cdot\,4......(n - 1)\,\cdot\,n}}$ =$\frac{1}{n}\,\cdot\,\frac{{n + 1}}{2}$ =$\mathop {Lim}\limits_{n \to \infty } \,\,\frac{{1 + \frac{1}{n}}}{2}$ =$\frac{1}{2}$
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