A Converging Sum

$\begin{aligned} S & = \lim_{n \to \infty} \sum_{i=1}^n \frac {i\cos \frac in - n \sin \frac in}{i^2} & \small \color{#3D99F6} \text{Divide up and down by }n \\ & = \lim_{n \to \infty} \sum_{i=1}^n \frac {\frac in \cos \frac in - \sin \frac in}{n \left(\frac in\right)^2} & \small \color{#3D99F6} \text{By Riemann sums: } \\ & = \int_0^1 \frac {x\cos x - \sin x}{x^2} dx & \small \color{#3D99F6} \lim_{n \to \infty} \sum_{k=a}^b \frac {f \left(\frac kn\right)}n = \int_\frac an^\frac bn f(x) \ dx \\ & = \frac {\sin x}x \ \bigg|_0^1 & \small \color{#3D99F6} \text{Since } \frac d{dx} \left(\frac {\sin x}x\right) = \frac {x\cos x - \sin x}{x^2} \\ & = \sin 1 - 1 & \small \color{#3D99F6} \text{Note that } \lim_{x \to 0} \frac {\sin x}x = 1 \end{aligned}$

Therefore, $10m+n = 10(1)+1 = \boxed{11}$ .

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Reference: Riemann sums

$S$ can be expressed as $\lim_{n→\infty}\sum_{i=1}^{n}\frac{\cos\frac{i}{n}}{i}-\frac{n\sin\frac{i}{n}}{i^{2}}=\lim_{n→\infty}\sum_{i=1}^{n}\frac{1}{n}\frac{\cos\frac{i}{n}}{\frac{i}{n}}-\frac{\sin\frac{i}{n}}{(\frac{i}{n})^{2}}$ .

Notice that this is a Riemann Sum of $\int_{0}^{1} \cos x-\frac{\sin x}{x}$ .

Then, using integration by parts, we can get $S=[\frac{\sin x}{x}]_{0}^{1}=\frac{\sin 1}{1}-\lim_{x→0}\frac{\sin x}{x}=\sin 1-1$ .

This gives the answer of $\boxed{11}$ .