If \(a=\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}\) and \(b=\lim _{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots+n^2}{n^3}\), then

\(a=\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2} \)
\( =\lim _{n \rightarrow \infty} \frac{n(n+1)}{2(n)(n)}=\lim _{n \rightarrow \infty}\left(\frac{1}{2}\right)\left(\frac{n+1}{n}\right)=\) \(\frac{1}{2} \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)=\frac{1}{2} \)
\( b=\lim _{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots .+n^2}{n^3} \)
\( =\lim _{n \rightarrow \infty} \frac{n(n+1)(2 n+1)}{6 n^3}=\lim _{n \rightarrow \infty} \frac{2 n^2+3 n+1}{6 n^2} \)
\( =\lim _{n \rightarrow \infty}\left(\frac{1}{6}\right)\left(2+\frac{3}{n}+\frac{1}{n^2}\right)=\frac{2}{6}=\frac{1}{3}\)
Thus \(\mathrm{a}=\frac{1}{2}\) and \(\mathrm{b}=\frac{1}{3} \Rightarrow 2 \mathrm{a}=3 \mathrm{~b}\)