Let $A_n$ be square matrix order 2 given $A_n=\left[\begin{array}{cc}{U}_{2n-1}& 0\\ 0& \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{U}_{2n}\end{array}\right], n\in N$ where $U_n=\underset{0}{\overset{\pi /2}{\int }}{\cos }^{n}x\cos nxdx,$ Then

$U_n=\underset{0}{\overset{\pi /2}{\int }}{\cos }^{n}x\cos nxdx=\frac{{\cos }^{n}x\sin nx}{n}\underset{0}{\overset{\pi /2}{\int }}-\underset{0}{\overset{\pi /2}{\int }}\frac{\sin nx}{n}n{\cos }^{n-1}x(-\sin x)$

$=\underset{0}{\overset{\pi /2}{\int }}\sin nx\sin x{\cos }^{n-1}xdx$

$\therefore 2U_n=U_{n-1}\Rightarrow U_n=\frac{U_{n-1}}{2}\Rightarrow U_n=\frac{\pi }{2^{n+1}}$

$t_r{A}_n=U_{2n-1}+U_{2n}=\frac{\pi }{2^{2n}}.\frac{3}{2}$

$\underset{n\to \infty }{Lt}\sum t_r{A}_n=\frac{3\pi }{2}\underset{n\to \infty }{lt}\sum \frac{1}{4^n}=\frac{3\pi }{2}\frac{1/4}{1-1/4}=\frac{\pi }{2}$