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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.

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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

$$ F_n(x) = \mathbb{E}\left[f\left(\frac{X_1+\cdots+X_n}{n}\right)\right] \xrightarrow[n\to\infty]{} f(\mathbb{E}[X_1]) = f(x). $$

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    $\begingroup$ 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 $a_n$. $\endgroup$ – Nilotpal Kanti Sinha Jul 27 at 10:58
  • $\begingroup$ Thank you for this super concise proof of that formula of Bernstein. $\endgroup$ – Giuseppe Negro Jul 29 at 17:13

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