 How Cheenta works to ensure student success?
Explore the Back-Story

# Test of Mathematics Solution Subjective 157 -Limit of a product  This is a Test of Mathematics Solution Subjective 157 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

## Problem

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 $\log (y) = \frac{1}{2n} { \log (1 + \frac{1}{2n}) + \cdots + \log (1 + \frac{2n}{2n}) }$ - $\frac{1}{2n} { \log (1 + \frac{2}{2n}) + \log (1 + \frac{4}{2n} )+ \log (1+ \frac{6}{2n}) + ... + \log (1+ \frac{2n}{2n}) }$

$\displaystyle \lim_{n \to \infty} \log (y) = \int_0^1 \log (1+x) dx - \frac{1}{2} \int_0^{1} \log(1+x) dx = \frac{1}{2} \int_{0}^1 \log (1+x) dx = \log 2 - \frac {1}{2}$

Therefore answer will be $e^{\log 2 - \frac {1}{2}}$ This is a Test of Mathematics Solution Subjective 157 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

## Problem

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 $\log (y) = \frac{1}{2n} { \log (1 + \frac{1}{2n}) + \cdots + \log (1 + \frac{2n}{2n}) }$ - $\frac{1}{2n} { \log (1 + \frac{2}{2n}) + \log (1 + \frac{4}{2n} )+ \log (1+ \frac{6}{2n}) + ... + \log (1+ \frac{2n}{2n}) }$

$\displaystyle \lim_{n \to \infty} \log (y) = \int_0^1 \log (1+x) dx - \frac{1}{2} \int_0^{1} \log(1+x) dx = \frac{1}{2} \int_{0}^1 \log (1+x) dx = \log 2 - \frac {1}{2}$

Therefore answer will be $e^{\log 2 - \frac {1}{2}}$

### Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy  