# Series - Bound by integrals

I have a series, $1^3 + 2^3 + 3^3 ... n^3$, and I want to find the upper and lower bound of this series using integrals. I know that for a series that is decreasing (such as $\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{2^2} ... \frac{1}{n^2}$), the bounds would be as follows:

$$f(x) = \frac{1}{x^2} \\ f(2) + f(3) + f(4)...f(n) \leq \int_1^n f(x)\,\mathrm{d}x \leq f(1) + f(2) + f(3)...f(n-1) \\ 1 \leq \sum_{n=0}^\infty \frac{1}{n^2} \leq 2$$

For a function that is increasing, like the original series, what would the bounds be? The same method wouldn't work because the function is increasing, but I can't seem to figure out the correct ones.

Thanks!

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The same method should work, but you just have to pay attention to the limits of integration. My usual way of figuring this out is drawing the value of the series as being the area under a bunch of rectangles of width 1 and height given by the value of each series term, and then trying to figure out what curve always lies above and below the tops of the rectangles. – Christopher A. Wong Sep 13 '12 at 1:02
I might note, however, that there is an explicit formula for the sum of cubes. – Christopher A. Wong Sep 13 '12 at 1:59

If you want to use integrals, the following will work. Draw the curve $y=x^3$, from $x=0$ to $x=n+1$. It need not be a good picture, as long as the curve you draw is concave up. It is enough to take say $n=4$ or $5$.
For $i=0$ to $n-1$, draw, in red, the rectangles with base $[i,i+1]$ and height $(i+1)^3$. The sum of the areas of the red rectangles is $1^3+2^3+\cdots +n^3$. It is easy to see from the picture that $$\int_0^n x^3\,dx \lt 1^3+2^3+\cdots+n^3\tag{1} .$$
Perhaps in a separate picture, for $i=1$ to $n$, draw, in blue, the rectangles with base $[i,i+1]$ and height $i^3$. The sum of the areas of the blue rectangles is $1^3+2^3+\cdots +n^3$. It is easy to see from the picture that $$1^3+2^3+\cdots+n^3 \lt \int_1^{n+1} x^3\,dx \tag{2}.$$ Calculate the integrals in $(1)$ and $(2)$ to find lower and upper bounds.
Remark: We can get better bounds by, for example, using the Trapezoidal Rule or the Midpoint Rule. As has been pointed out in a comment, our sum is exactly $\dfrac{n^2(n+1)^2}{4}$, so there is really no need for bounds. However, the same idea will work for $1^{2.5}+2^{2.5}+\cdots +n^{2.5}$.