Limit of a product (TOMATO Subjective 157)

Evaluate \(\mathbf { \lim_{n to \infty } { (1 + \frac{1}{2n}) (1 + \frac{3}{2n} )(1+ \frac{5}{2n}) + … + (1+ \frac{2n-1}{2n})}^{\frac{1}{2n}} }\)

Discussion:

Let \(\mathbf { y = { (1 + \frac{1}{2n}) (1 + \frac{3}{2n} )(1+ \frac{5}{2n}) + … + (1+ \frac{2n-1}{2n})}^{\frac{1}{2n}} }\)

Then \(\mathbf { \log (y) = \frac{1}{2n}{ \log (1 + \frac{1}{2n}) + \log (1 + \frac{3}{2n} )+ \log (1+ \frac{5}{2n}) + … + log (1+ \frac{2n-1}{2n})} }\)

This implies \(\log (y) = \frac{1}{2n} { \log (1 + \frac{1}{2n}) + \cdots + \log (1 + \frac{2n}{2n}) } \) – \(\frac{1}{2n} { \log (1 + \frac{2}{2n}) + \log (1 + \frac{4}{2n} )+ \log (1+ \frac{6}{2n}) + … + \log (1+ \frac{2n}{2n}) } \)

Hence \(\lim_{n \to \infty} \log (y) = \int_0^1 \log (1+x) dx – \frac{1}{2} \int_0^{1} \log(1+x) dx = \frac{1}{2} \int_{0}^1 \log (1+x) dx = \log 2 – \frac {1}{2} \) (Answer)

Leave a Reply

Your email address will not be published. Required fields are marked *