Let $f:[a,\infty)\rightarrow \mathbb{R}$ be a uniformly continuous function. $\int_{a}^{\infty} f$ converges.Prove that $\lim_{x\to\infty} f(x)=0$ 
Let $f:[a,\infty)\rightarrow \mathbb{R}$ be a uniformly continuous function in that range. $\int_{a}^{\infty} f$ converges. Prove that $\lim_{x\to\infty} f(x)=0$
Hint: Use the sequence $F_n(x)=n\int_{x}^{x+\frac{1}{n}} f$.

Honestly I have been trying to solve this one for some time but the hint really confuses me.
I have tried to mess around with $F_n(x)$ a bit, for example by using the fundamental theorem but it still seems like such a random choice and I can't make anything out of it.
Any guidance/explanations will be appreciated.
Please use the hint in the question.

Assume that $\lim_{x \to \infty}f(x) =0$ does not hold and arrive  at contradiction with the fact that the integral of $f$ is convergent.
If $\lim_{x \to \infty} f(x) = 0$ does not hold then there exists $\epsilon_0 > 0$ and a sequence $x_n \to \infty$ such that $|f(x_n)| \geqslant \epsilon_0$ for all $n$. 
Assume WLOG that $f(x_n) \geqslant \epsilon_0$.  
There exists by uniform continuity of $f$ a $\delta > 0$ such that $|f(t) - f(x_n)| < \epsilon_0/2$ for all $t \in [x_n - \delta,x_n + \delta].$ 
This implies $f(t) > \epsilon_0/2$ and
$$ \int_{x_n - \delta}^{x_n + \delta} f(t) \, dt > \epsilon_0\delta$$
This violates the Cauchy criterion for convergence of the improper integral since $x_n$ can be arbitrarily large.
