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

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Aicharya · Accepted Answer

$\begin{aligned} & a=\lim _{n ightarrow \infty} \frac{1+2+3+\ldots+n}{n^2} \ & =\lim _{n ightarrow \infty} \frac{n(n+1)}{2(n)(n)}=\lim _{n ightarrow \infty}\left(\frac{1}{2}ight)\left(\frac{n+1}{n}ight)=\frac{1}{2} \lim _{n ightarrow \infty}\left(1+\frac{1}{n}ight)=\frac{1}{2} \ & b=\lim _{n ightarrow \infty} \frac{1^2+2^2+3^2+\ldots .+n^2}{n^3} \ & =\lim _{n ightarrow \infty} \frac{n(n+1)(2 n+1)}{6 n^3}=\lim _{n ightarrow \infty} \frac{2 n^2+3 n+1}{6 n^2} \ & =\lim _{n ightarrow \infty}\left(\frac{1}{6}ight)\left(2+\frac{3}{n}+\frac{1}{n^2}ight)=\frac{2}{6}=\frac{1}{3} \end{aligned}$ Thus $\mathrm{a}=\frac{1}{2}$ and $\mathrm{b}=\frac{1}{3} \Rightarrow 2 \mathrm{a}=3 \mathrm{~b}$

If $a=\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}$ and $b=\lim _{n \rightarrow \infty}…

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