Studies and fun

TODAY'S PROBLEM

Evaluate $\mathbf { limit_{n \to \infty } \{ (1 + \frac{1}{2n}) (1 + \frac{3}{2n} )(1+ \frac{5}{2n}) + ... + (1+ \frac{2n-1}{2n})\}^{\frac{1}{2n}} }$

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 $\mathbf { log (y) = \frac{1}{2n}\{ log (1 + \frac{1}{2n}) + log (1 + \frac{2}{2n}) +log (1 + \frac{3}{2n} )+ log (1 + \frac{4}{2n})+ log (1+ \frac{5}{2n}) + ... + log (1+ \frac{2n-1}{2n}) + log (1 + \frac{2n}{2n})\} }$ – $\mathbf { \frac{1}{2n}\{ log (1 + \frac{2}{2n}) + log (1 + \frac{4}{2n} )+ log (1+ \frac{6}{2n}) + ... + log (1+ \frac{2n}{2n})\} }$

Hence $\mathbf { limit_{n \to \infty} log (y) = \int_0^1 log (1+x) dx - \int_0^{1/2} log(1+x) dx = \int_{1/2}^1 log (1+x) dx = -\frac{1}{2} + \frac{1}{2} log {\frac{3}{2}} }$ (Answer)

Studies and fun

Friday, 25 July 2014

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