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!
 A: 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 $1^{2.5}+2^{2.5}+\cdots +n^{2.5}$.
