Ask Mathematics

Is this sequence $x_n=(1-\frac12)^{(\frac12-\frac13)^{…^{(\frac{1}{n}-\frac{1}{n+1})}}}$ uniformly distributed modulo 1?

The sequence $(x_n) ,n=1,2,\cdots $ is uniformly distributed mod 1 iff for real -valued continious function $f$ defined on the closed interval $[0,1]$ we have $ \lim_{N\to \infty} \frac1N \sum_{n=1}^{N} f(\{x_n\})=\int_{0}^{1} f(x) dx$, Really am not able to apply that theorem for the following sequence $x_n=(1-\frac12)^{(\frac12-\frac13)^{…^{(\frac{1}{n}-\frac{1}{n+1})}}}$ to know if it is uniformly distributed or not ? However that theorem dosn’t need the convergence of a such sequence because the titled sequence converge according to the parity of $n$ which lie in $[0,1]$

Leave a Reply

Your email address will not be published.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