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Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a continuous and periodic function with period $P$ and let $F$ be a primitive of $f$.
Prove that: $$\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{k=1}^{n}\frac{F(k)}{k}\cdot f\left ( \frac{k}{n} \right )=\frac{1}{P}\cdot \int_{0}^{P}f(x)dx\cdot \int_{0}^{1}f(x)dx$$

I know that $\frac{1}{P}\cdot \int_{0}^{P}f(x)dx=f(c)$ where $c\in (0,P)$ and that the LHS of the equation could be written as a Riemann sum, but I haven't found something helpful yet.

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    Its clear that the L.H.S is in some form of the Riemman integral $ \lim_{n \rightarrow \infty} \frac{b-a}{n}\sum_{k=1}^{n} f\left(\frac{k(b-a)}{n}\right)$ where a = $0$ and b = $1$ which is the second integral on the R.H.S $\int_0^1 f(x)dx$. Now you just need to show that $\frac{F(k)}{k}$ in the sum is equal to the mean value over the period as you've shown in the first integral on the RHS as $f(c)$ – 2017-02-19
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    But is your statement true? Because $ \sum_{k=1}^{n}\frac{F(k)}{n}\cdot f\left ( \frac{k}{n} \right ) $ is not equal with $\sum_{k=1}^{n}\frac{F(k)}{n}\cdot \sum_{k=1}^{n} f\left ( \frac{k}{n} \right )$. – 2017-02-20
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    No, I'm not saying that you can separate them like that, as summations do not have that property generally. I'm suggesting you can make some transformations to variables and transform $\frac{F(k)}{k}$ into $\frac{1}{P} \int_0^P f(x)dx$, which means it is a constant and doesn't rely on $k$ so that you can pull it out of the summation. Knowing that the function is periodic might help you find that transformation. – 2017-02-20
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    For example you know that $F(k) = \int_0^k f(x) dx$ by the fundamental theorem of calculus – 2017-02-20
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    Well, the only relation I've found is that $F(x+P)=F(x)+c$, where $c$ is a constant...I got stuck there. – 2017-02-20
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    After some obvious simplification the problem boils down to showing that $$\lim_{n \to \infty}\frac{1}{n}\sum_{k = 1}^{n}\frac{1}{k}f\left(\frac{k}{n}\right) = 0$$ – 2017-02-21

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