Recognizing a Riemann Sum to Compute a Limit

The sum–integral link can also be read backwards: some limits of sums, impossible to compute directly, turn out to be integrals in disguise. It is a powerful tool.

Take the most common case, on the interval $[0 \,;\, 1]$. Here $a = 0$ and $b = 1$, so $\Delta x = \tfrac{1}{n}$ and the points are $x_k = \tfrac{k}{n}$. The definition becomes:

$$\lim_{n \to +\infty} \frac{1}{n}\sum_{k=1}^{n} f\!\left(\frac{k}{n}\right) = \int_0^1 f(x)\,\mathrm{d}x$$

In other words: as soon as a sum has the form "$\tfrac{1}{n}$ times a sum of $f\!\left(\tfrac{k}{n}\right)$", its limit is simply the integral of $f$ over $[0 \,;\, 1]$.

Example. What is the limit of the following sum?

$$\frac{1}{n}\left[\left(\frac{1}{n}\right)^2 + \left(\frac{2}{n}\right)^2 + \cdots + \left(\frac{n}{n}\right)^2\right] = \frac{1}{n}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^2$$

We recognize the form, with $f(x) = x^2$. The limit is therefore:

$$\int_0^1 x^2\,\mathrm{d}x = \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3}$$

A sum that seemed endless is computed in one stroke, by reading it as an area. Remember the method: spot the factor $\tfrac{1}{n}$, identify $f\!\left(\tfrac{k}{n}\right)$, then replace the limit with the integral over $[0 \,;\, 1]$. It is one of the most beautiful bridges between the discrete and the continuous.