## Solution

In the Warm-up we saw that $n^2$ and $n^3$ can be used as lower and upper bounds for the value of $\displaystyle\sum_{r=1}^{n} r^2$: $n^2 ≤ \displaystyle\sum_{r=1}^{n} r^2 ≤ n^3$

We can use the diagrams below to find a different way to bound $\displaystyle\sum_{r=1}^{n} r^2$.

• Which of the areas represented in the two diagrams is bigger?
• Does this result generalise?

The first diagram is a representation of the sum of the first $8$ square numbers, where each bar has width $1$ and height $r^2$. The second diagram has a shaded area which can represented by the integral $\int_0^8 x^2 \, dx$.

Which of these areas is bigger? We could numerically compute the areas but a simple way to compare them is to draw the two images on the same axes.

We can see that $\int_0^8 x^2 \, dx < \sum_{r=1}^{8} r^2$, but we can also generalise this result from the diagram and say that

$\int_0^n x^2 \, dx < \sum_{r=1}^{n} r^2.$

Use this information to provide either an upper or lower bound for the value of $\displaystyle\sum_{r=1}^{n} r^2$.

Calculating the value of the integral, we have $\int_0^n x^2 \, dx = \frac{1}{3}n^3$, so we can provide a lower bound for $\sum_{r=1}^{n} r^2$,

$\dfrac{1}{3}n^3 < \sum_{r=1}^n r^2.$

How does what we’ve done here, using an integral to estimate a summation, compare to what you’ve done before using integrals and sums?

Can you find a function that bounds the graphical representation of $\displaystyle\sum_{r=1}^{n} r^2$ on the other side?

Use this function to provide a second bound for $\displaystyle\sum_{r=1}^{n} r^2$.

Now we know that we are looking to find an upper bound. We can find a function that gives an overestimate of the area by using a transformation of $y = x^2$.

Therefore the integral $\int_0^n (x+1)^2 \, dx$ provides an upper bound to $\sum_{r=1}^{n} r^2$. It evaluates to $\int_0^n (x+1)^2 \, dx = \frac{1}{3}n^3 + n^2 + n$, so we have the following bounds for $\sum_1^n r^2$:

$\dfrac{1}{3}n^3 < \sum_1^n r^2 < \dfrac{1}{3}n^3 + n^2 + n.$

Our result shows that $\sum_1^n r^2 \approx \frac{1}{3}n^3$. Does this seem reasonable? Why might we have expected this result?

There is more than one way we can approximate an upper bound for $\sum_1^n r^2$. A different approach involves the following diagram.

We can use this to write the inequality $\displaystyle\sum_0^{n-1} r^2 < \dfrac{1}{3}n^3.$

If we add $n^2$ to both sides of the inequality we get $\displaystyle\sum_0^{n-1} r^2 + n^2 < \dfrac{1}{3}n^3 + n^2,$

which can be simplified to

$\displaystyle\sum_0^{n} r^2 < \dfrac{1}{3}n^3 + n^2.$

How does this compare to the upper bound found above? Does it matter that the limits are slightly different?

To find the exact value of $\sum_1^n r^2$ you might like to try the review question R6143.