Posts

Showing posts from February, 2020

A problem proposed by prof. Babev

Consider the sequence a_n=\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t. It could be proven that \lim_{n\to \infty}n a_n=\frac{1}{2}  Now we are going to prove that \lim_{n\to\infty}n\left(na_n-\frac{1}{2}\right)=-\frac{1}{4} and show how to derive all such limits. \textbf{Proof.} Since \displaystyle\int_{n}^{n+1}\sin^2(\pi t)\text{d}{t}=\frac{1}{2} (for any n\in \mathbb{N}) we can rewrite the limit in question as \lim_{n\to \infty}n\left(n\left(\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t-\int_{n}^{n+1}\frac{\sin^2(\pi t)}{n}\text{d}t\right)\right)=\lim_{n\to \infty}n^2\left(\int_{n}^{n+1}\sin^2(\pi t)\left(\frac{1}{t}-\frac{1}{n}\right)\text{d}t\right) =\lim_{n\to \infty}n\left(\int_{n}^{n+1}\sin^2(\pi t)\frac{n-t}{t}\text{d}t\right) Changing the variables t\to n+s and using periodicity of sine, the last simplifies to \lim_{n\to \infty}n\int_{0}^{1}\sin^2(\pi s)\frac{-s}{n+s}\text{d}s Since for s\in[0,1], $\displaystyle \...