Evaluate $\lim_{n\to\infty} \sum_{i=1}^{n} \left[\frac{4}{n} - \frac{i^2}{n^3}\right]$ $$\lim_{n\to\infty} \sum_{i=1}^{n} \left[\frac{4}{n} - \frac{i^2}{n^3}\right]$$
If I take out $\frac{1}{n}$, it looks like a Riemann sum:
$$\lim_{n\to\infty} \sum_{i=1}^{n} \left[\left(\frac{1}{n}\right)\left(4 - \frac{i^2}{n^2}\right)\right]$$
How should I proceed with solving this limit? The given answer is $\frac{11}{3}$.
 A: A different, possibly more basic (depending on how you define "basic") solution is to use the sum for the squares of the first $n$ integers (as suggested by RRL in the question comments) being
$$\sum_{i=1}^n i^2 = \cfrac{n\left(n+1\right)\left(2n + 1\right)}{6} = \cfrac{2n^3 + 3n^2 + n}{6} \tag{1}\label{eq1}$$
Thus,
\begin{align}
\lim_{n \to \infty} \sum_{i=1}^n \left[\cfrac{4}{n} - \cfrac{i^2}{n^3}\right] & = \lim_{n \to \infty} \left[\cfrac{4n}{n} - \cfrac{2n^3 + 3n^2 + n}{6n^3} \right] \\
& = \lim_{n \to \infty} \left[4 - \cfrac{1}{3} - \cfrac{1}{2n} - \cfrac{1}{6n^2} \right] \\
& = \cfrac{11}{3} \tag{2}\label{eq2}
\end{align}
A: You're almost there. Taking the limit of the Riemann sum leads to
\begin{align}\lim_{n\to\infty} \sum_{i=1}^{n} \left[\left(\frac{1}{n}\right)\left(4 - \frac{i^2}{n^2}\right)\right]
&=\lim_{n\to\infty} \sum_{i=1}^{n} \left[\left(\frac{i+1}{n}-\frac{i}{n}\right)\left(4 - \left(\frac{i}{n}\right)^2\right)\right] \\
&=\int_0^1(4-x^2)dx
\end{align}
and this is indeed $\frac{11}{3}$.
