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.
 A: 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: 
$$ \sum_{i=1}^{n}i^2 = \frac{n\cdot (n+1) \cdot 2n+1)}{6}$$

$$\begin{split}
\lim_{n \to \infty }\frac{981}{n+5}\sum_{i=1}^{n} \left(\frac{i^2}{n^2} \right)
&= \lim_{n \to \infty} \frac{981}{n+5} \frac{\sum_{i=1}^{n}i^2}{n^2}\\
&= \lim_{n \to \infty}\frac{981}{n+5} \frac{n(n+1)(2n+1)}{6 \cdot n^2}\\
&= \lim_{n \to \infty} \frac{n}{n} \cdot \frac{981 (2n^2+3n+1)}{(6n^2 +30n)}\\
&= \lim_{n \to \infty}\frac{n^2}{n^2} \cdot \frac{981\cdot(2+\frac{3}{n} + \frac{1}{n^2})}{6(1 + \frac{30}{n})} \\
&= \lim_{n \to \infty}\frac{327\cdot(2+\frac{3}{n} + \frac{1}{n^2})}{2(1 + \frac{30}{n})}  &= 327
\end{split}$$
A: Firstly calculate the sum$$\sum_{i=1}^{n}\frac{i^2}{n^2}=\frac{1}{n^2}\sum_{i=1}^{n}i^2=\frac{1}{n^2}\frac{n(n+1)(2n+1)}{6}=\frac{\not n^2}{\not{n^2}}\frac{(1+\frac1n)(2n+1)}{6}$$ Therefore $$\begin{align*}\lim_{n \to \infty}\frac{981}{n+5}\frac{(1+\frac1n)(2n+1)}{6}&=\lim_{n\to \infty}\frac{\not n}{\not n}\frac{981(1+\frac1n)(2+\frac1n)}{6(1+\frac5n)}\overset{\frac1n\to 0}=\\\\&=\frac{981(1+0)\cdot(2+0)}{6(1+0)}=\frac{981\cdot2}{6}=\frac{981}{3}=327\end{align*}$$
A: Hint: Pull the $\frac{1}{n^2}$ out of the sum and use the identity $$\sum_{i=1}^ni^2= \frac{n(n+1)(2n+1)}{6}$$ 
A: I see a (calculus) tag.  So ... that homework problem must follow a chapter on Riemann sums ...  Let's try that.
Consider the integral
$$
\int_0^1 x^2\,dx = \frac{1}{3} .
$$
That integrand $x^2$ is continuous, therefore Riemann integrable.  Consider the "upper sums" for $n$ equal-size intervals.
$$
\sum_{i=1}^n \left(\frac{i}{n}\right)^2\cdot\frac{1}{n} \to \frac{1}{3}
$$
as $n \to \infty$.  So the original probem must have answer $981/3 = 327$.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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With Stolz-Ces$\grave{a}$ro Theorem:
\begin{align}
\color{#f00}{\lim_{n \to \infty}{981 \over n + 5}
\sum_{i = 1}^{n}\pars{i^{2} \over n^{2}}} & =
981\lim_{n \to \infty}{1 \over \pars{n + 6}\pars{n + 1}^{2} - \pars{n + 5}n^{2}}
\pars{\sum_{i = 1}^{n + 1}i^{2} - \sum_{i = 1}^{n}i^{2}}
\\[3mm] &=
981\lim_{n \to \infty}{n^{2} + 2n + 1 \over 3n^{2} + 13n + 6} =
327\lim_{n \to \infty}{1 + 2/n + 1/n^{2} \over 1 + 13/\pars{3n} + 2/n^{2}} =
\color{#f00}{327}
\end{align}
