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$$\lim_{n\to\infty} \sum_{i=1}^{n} \left[\frac{4}{n} - \frac{i^2}{n^3}\right]$$

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 $\frac{11}{3}$.

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    $\begingroup$ You suggested a Riemann sum. What is $\int_0^1 (4 - x^2) \, dx$? $\endgroup$ – RRL Feb 7 at 1:30
  • $\begingroup$ That evaluates to $11/3$, but can you explain how you got that? $\endgroup$ – Infuzion Feb 7 at 1:31
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    $\begingroup$ $\frac{1}{n} \sum_{j=1}^n f(j/n) \to \int_0^1 f(x) \, dx$. Also you can work with $\sum_{j=1}^n j^2 $ which has a closed form. $\endgroup$ – RRL Feb 7 at 1:33
  • $\begingroup$ Thanks. That makes sense now $\endgroup$ – Infuzion Feb 7 at 1:41
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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 $\frac{11}{3}$.

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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}

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