limit ( 3 )

Calculus Level 4

$\lim_{n\to \infty} \bigg[ \prod_{k=0}^{n-1} cos\bigg( \frac{2^{k}. \pi}{2^{n}-1} \bigg) \bigg] = ?$

The answer is 0.

1 solution

Refaat M. Sayed
Jul 25, 2015

$L= \lim_{n\to \infty} \bigg[ \prod_{k=0}^{n-1} cos\bigg( \frac{2^{k}. \pi}{2^{n}-1} \bigg) \bigg]$ $L=\lim_{n\to \infty} \bigg( cos\bigg( \frac{\pi}{2^{n}-1} \bigg) cos\bigg( \frac{2\pi}{2^{n}-1} \bigg)......cos\bigg( \frac{\pi}{2}.\frac{2^n}{2^{n}-1} \bigg)$ $\lim_{n\to \infty} \frac{2^n}{2^{n}-1} = \lim_{n\to \infty} \frac{1}{1-2^{-n}} = 1$ $\text{so} \lim_{n\to\infty} cos \bigg( \frac{\pi}{2} . \frac {2^n}{2^{n}-1} \bigg) = cos\bigg( \frac{\pi}{2} \bigg) = 0$ $L = \lim_{n\to \infty} \bigg[ \prod_{k=0}^{n-1} cos\bigg( \frac{2^{k}. \pi}{2^{n}-1} \bigg) \bigg] = 0$

Plugging in the limiting value of a particular part of the function is not always correct. Had it been some other term in the product (not in this case though) which has a tendency to approach infinity , you couldn't have claimed that the limit of the entire product is 0.

Ankit Kumar Jain - 3 years, 1 month ago

limit ( 3 )

The answer is 0.

1 solution

0 pending reports