If a = lim_{n rightarrow infty} frac{1+2+3+ld | vedclass

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(B) First,we evaluate $a$: $a = \lim_{n \rightarrow \infty} \frac{n(n+1)}{2n^2} = \lim_{n \rightarrow \infty} \frac{n^2+n}{2n^2} = \lim_{n \rightarrow

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$2a = 3b$

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(B) First,we evaluate $a$: $a = \lim_{n \rightarrow \infty} \frac{n(n+1)}{2n^2} = \lim_{n \rightarrow \infty} \frac{n^2+n}{2n^2} = \lim_{n \rightarrow \infty} \frac{1 + \frac{1}{n}}{2} = \frac{1}{2}$. Next,we evaluate $b$: $b = \lim_{n \rightarrow \infty} \frac{n(n+1)(2n+1)}{6n^3} = \lim_{n \rightarrow \infty} \frac{2n^3 + 3n^2 + n}{6n^3} = \lim_{n \rightarrow \infty} \frac{2 + \frac{3}{n} + \frac{1}{n^2}}{6} = \frac{2}{6} = \frac{1}{3}$. Comparing the values,$a = \frac{1}{2}$ and $b = \frac{1}{3}$. Therefore,$2a = 2(\frac{1}{2}) = 1$ and $3b = 3(\frac{1}{3}) = 1$. Thus,$2a = 3b$.

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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