Limits

Calculus Level 4

Evaluate

$\lim_{n \rightarrow \infty} \prod_{i=3}^n \frac{ i^2 - 4 } { i^2}.$

The answer is 0.166667.

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Wei Xian Lim
Feb 7, 2015

$\prod_{i=3}^{\infty}\frac{i^2-4}{i^2}= \prod_{i=3}^{\infty}(\frac{i-2}{i})(\frac{i+2}{i})= \prod_{i=3}^{\infty}\frac{i-2}{i}\prod_{i=3}^{\infty}\frac{i+2}{i}$ This can be written as: $\bigg[\bigg(\frac{1}{3}\bigg)\bigg(\frac{2}{4}\bigg)\bigg(\frac{3}{5}\bigg)\bigg(\frac{4}{6}\bigg)...\;...\bigg] \bigg[\bigg(\frac{5}{3}\bigg)\bigg(\frac{6}{4}\bigg)\bigg(\frac{7}{5}\bigg)\bigg(\frac{8}{6}\bigg)...\;...\bigg]$ Cancelling out intermediate terms, we obtain: $\frac{1\cdot2}{3\cdot4}=\LARGE\boxed{\frac{1}{6}}$

The last step can also be done formally like this:

$\left(\prod_{i=3}^\infty \frac{i-2}{i}\right)\left(\prod_{i=5}^\infty \frac{i}{i-2}\right)\\ = \frac{1}{3}\times \frac{2}{4}\times \prod_{i=5}^\infty\left(\frac{i-2}{i}\cdot \frac{i}{i-2}\right)\\= \frac{1}{6}\times \prod_{i=5}^\infty (1) = \boxed{\frac{1}{6}}$

Prasun Biswas - 6 years, 4 months ago

Nice Observation !!

A Former Brilliant Member - 6 years, 4 months ago

Are you implying Wei Xian's last step wasn't done formally? Sigma/pi form does not automatically mean that your steps are rigorous.

Jake Lai - 6 years, 3 months ago

Nope, I'm implying that I'm lazy. :P

Prasun Biswas - 6 years, 3 months ago

Limits

The answer is 0.166667.

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