# Binomial analogue of Riemann sum for definite integral

I found this interesting relationship which is an analogue of the Riemann sum for definite integral.

$$\lim_{n \to \infty}\frac{1}{n}\sum_{r = 1}^n f\Big(\frac{r}{n}\Big) = \int_{0}^{1}f(x) dx$$

$$\lim_{n \to \infty}\frac{1}{2^n}\sum_{r = 1}^n {n \choose r}f\Big(\frac{r}{n}\Big) = f\Big(\frac{1}{2}\Big)$$

Application. $$\lim_{n \to \infty}\frac{1}{2^n}\sum_{r = 1}^n {n \choose r}\Gamma\Big(\frac{r}{n}\Big) = \sqrt{\pi}$$

$$\lim_{n \to \infty}\frac{1}{2^n}\sum_{r = 1}^n {n \choose r}\arcsin\Big(\frac{r}{n}\Big) = \frac{\pi}{6}$$

Question: Any reference to this in mathematics literature?

A generalization of this problem is posted here.

For a continuous function $$f:[0,1]\to\mathbb{R}$$, it is well-known that $$f$$ can be uniformly approximated by a linear combination of Bernstein polynomials, i.e., the function
$$F_n(x) = \sum_{k=0}^{n} \binom{n}{k}x^k(1-x)^{n-k}f\left(\frac{k}{n}\right)$$
converges uniformly to $$f$$ on $$[0, 1]$$ as $$n\to\infty$$. Your result is a special case with $$x = \frac{1}{2}$$. The proof can be found in many analysis book, such as Marsden's Elementary Classical Analysis as well as in the wiki article linked above.
For each fixed $$x \in [0, 1]$$, this can also be understood as an application of the strong law of large numbers, i.e., if $$(X_i)_{i\geq 1}$$ is a sequence of i.i.d. $$\operatorname{Bernoulli}(x)$$ RVs, then
$$F_n(x) = \mathbb{E}\left[f\left(\frac{X_1+\cdots+X_n}{n}\right)\right] \xrightarrow[n\to\infty]{} f(\mathbb{E}[X_1]) = f(x).$$
• Very precise an to the point answer I have accepted it. While going through the proof in the wiki link, I saw that $K/n$ originates form the weak law of large numbers i.e. getting $k$ successes from $n$ trials. Is there a variation of this result where $f(k/n)$ can be replaced $f(a_k/a_n)$ for some suitable sequence $a_n$. – Nilotpal Kanti Sinha Jul 27 at 10:58