<s>[INST] Compute $\lim\limits_{n \to \infty }\frac{981}{n+5}\sum_{i=1}^{n} \left (\frac{i^2}{n^2} \right)$ Compute the given limit
$$
\lim_{n \to \infty }\frac{981}{n+5}\sum_{i=1}^{n} \left (\frac{i^2}{n^2} \right)
$$
The sum is: 
Can someone please show me the steps to complete this problem?  The answer I arrived at was 0 but the homework program is telling me that it's wrong.
Thank you.
 [/INST] Hint1:
$$\begin{split}
\lim_{n \to \infty }\frac{981}{n+5}\sum_{i=1}^{n} \left(\frac{i^2}{n^2} \right)
&= \lim_{n \to \infty} \frac{981}{n+5} \cdot \frac{\sum_{i=1}^{n}i^2}{n^2}\\
\end{split}$$
Hint2: 
$$ \sum_{i=1}^{n}i^2 = \frac{n\cdot (n+1) \cdot 2n+1)}{6}$$

$$\begin{split}
\lim_{n \to \infty }\frac{981}{n+5}\sum_{i=1}^{n} \left(\frac{i^2}{n^2} \right)
&= \lim_{n \to \infty} \frac{981}{n+5} \frac{\sum_{i=1}^{n}i^2}{n^2}\\
&= \lim_{n \to \infty}\frac{981}{n+5} \frac{n(n+1)(2n+1)}{6 \cdot n^2}\\
&= \lim_{n \to \infty} \frac{n}{n} \cdot \frac{981 (2n^2+3n+1)}{(6n^2 +30n)}\\
&= \lim_{n \to \infty}\frac{n^2}{n^2} \cdot \frac{981\cdot(2+\frac{3}{n} + \frac{1}{n^2})}{6(1 + \frac{30}{n})} \\
&= \lim_{n \to \infty}\frac{327\cdot(2+\frac{3}{n} + \frac{1}{n^2})}{2(1 + \frac{30}{n})}  &= 327
\end{split}$$
</s>