Interesting Limit (3)

If $f$ is continuously differentiable on $[0,1]$ , and if $0 \le a < b \le 1$ , then $\begin{aligned} \tfrac12(b-a)[f(a) + f(b)] - \int_a^b f(t)\,dt & = \; \tfrac12\int_a^b \big[ f(a) + f(b) - 2f(t)\big]\,dt \; = \; \tfrac12\int_a^b \left\{ \int_t^b f'(u)\,du - \int_a^t f'(u)\,du \right\}\,dt \\ & = \; \tfrac12\int_a^b \big[(u-a) - (b-u)\big] f'(u)\,du \; = \; \int_a^b \big[u - \tfrac12(a+b)\big]f'(u)\,du \; = \; \int_a^b \big[u - \tfrac12(a+b)][f'(u) - f'(a)]\,du \end{aligned}$ and hence $\left|\tfrac12(b-a)[f(a) + f(b)] - \int_a^b f(t)\,dt \right| \; \le \; \tfrac14K_{b-a}(b-a)^2$ where $K_\delta = \mathrm{sup}\big\{ |f'(u) - f'(v)| \; : \; 0 \le u,v \le 1\,,\,|u-v| \le \delta \big\}$ . Thus we deduce that $\left| \tfrac{1}{2n} \big[f(\tfrac{k}{n}) + f(\tfrac{k+1}{n})\big] - \int_{\frac{k}{n}}^{\frac{k+1}{n}} f(t)\,dt\right| \; \le \; \frac{K_{n^{-1}}}{4n^2}$ for any positive integer $n$ and $0 \le k \le n-1$ . Adding these inequalities up, we see that $\left| n^{-1}\left(\tfrac12f(0) + \sum_{k=1}^{n-1}f(\tfrac{k}{n}) + \tfrac12f(1)\right) - \int_0^1f(t)\,dt\right| \; \le \; \frac{K_{n^{-1}}}{4n}$ and hence $\left| \sum_{k=1}^n f(\tfrac{k}{n}) - n\int_0^1 f(t)\,dt - \tfrac12[f(1) - f(0)]\right| \; \le \; \tfrac14K_{n^{-1}}$ Since $f$ is continuously differentiable, $K_\delta \to 0$ as $\delta \to 0$ , and hence $\lim_{n \to \infty} \left\{ \sum_{k=1}^n f(\tfrac{k}{n}) - n\int_0^1 f(t)\,dt \right\} \; = \; \tfrac12[f(1) - f(0)]$ In this case, we need $f(t) = \tan^{-1}t$ , and hence $\begin{aligned} A & = \; \tfrac12[f(1) - f(0)] \; = \; \tfrac18\pi \\ B & = \; \int_0^1 f(t)\,dt \; = \; \Big[t \tan^{-1}t - \tfrac12\ln(t^2 + 1)\Big]_0^1 \; = \; \tfrac14\pi - \tfrac12\ln2 \end{aligned}$ so that $4A - 2B = \ln2$ , making the answer $\boxed{2}$ .

Relevant wiki: Taylor's Theorem (with Lagrange Remainder)

In order for the limit to exist, we must have $B = \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\tan^{-1}\left(\frac{k}{n}\right) = \int_0^1 \tan^{-1}(x)\,dx = x\tan^{-1}(x) - \frac{1}{2}\ln(x^2+1)\Big|_0^1 = \frac{\pi}{4} - \frac{1}{2}\ln 2$

Then, using the integral form for $B$ and writing $f(x) = \tan^{-1}(x)$ , we have $\begin{aligned} \sum_{k=1}^n \tan^{-1}\left(\frac{k}{n}\right) - Bn &= n\left(\frac{1}{n}\sum_{k=1}^n \tan^{-1}\left(\frac{k}{n}\right) - \int_0^1 \tan^{-1}\,dx\right) \\ &= n\sum_{k=1}^n \int_{(k-1)/n}^{k/n} \left(f\left(\frac{k}{n}\right) - f(x)\right)\,dx \\ &= n\sum_{k=1}^n \frac{1}{2}f'(x_{k,n})\cdot \frac{1}{n^2} \quad \text{ for some } x_{k,n} \in \left(\frac{k-1}{n},\frac{k}{n}\right) & \textbf{[Taylor's theorem with Lagrange Remainder]} \\ &= \frac{1}{2} \cdot \frac{1}{n} \sum_{k=1}^n f'(x_{k,n}) \\ &\xrightarrow{n\to\infty} \frac{1}{2} \int_0^1 f'(x)\,dx = \frac{1}{2} [f(1) - f(0)] = \frac{\pi}{8} := A \end{aligned}$

Therefore, $e^{4A-2B} = e^{\ln 2} = \boxed{2}$