Somemathhere

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 \...