$$P=\sum_{r=3n}^{4n-1} \frac{r^2+13n^2-7rn}{n^3}$$. $$Q=\sum_{r=3n+1}^{4n} \frac{r^2+13n^2-7rn}{n^3}$$. $$I=\int_{3}^{4} (x^2-7x+13) dx = \frac{5}{6}$$ Compare the values of $P,Q,I$
I know that for a strictly increasing function $\sum_{r=0}^{n-1} f(x) \leq \int_{0}^{1} f(x)dx \leq \sum_{r=1}^{n} f(x)dx$ and reverse for a strictly decreasing function. (I am also comfortable with changing the lower/upper limits)
However, the function here, $x^2-7x+13$ is strictly decreasing from $x=3$ to $x=3.5$ and strictly increasing from $x=3.5$ to $x=4$.
I'm unable to compare the magnitude of "gain" in one interval with "loss" in the other to decide the overall outcome of the inequality.
I thought that the concavity of the function is $2$ so the derivative is increasing, and hence the latter parts change at a faster rate, hence the interval $3.5$ to $4$ outways the effect of the interval $3$ to $3.5$.
So I think that $P<I$ and $Q>I$. Is this correct?
How to solve this question rigorously?
I'm going to assume $n$ is even for simplicity.
First, notice that you have a symmetric parabola centered at $x=3.5$. So by symmetry $P=Q$.
Let us only think about the definition of $P$. If instead of a parabolic $f$ we had a symmetric piecewise linear function (decreasing until $3.5$ and increasing after, see the picture below), the difference between the sum and integral from increasing and decreasing parts would exactly cancel out:
But for the parabola, because of its convexity, the overestimation from the left side is larger than the underestimation from the right side:
You can write a formal proof for this using the definition of convexity.
So the final answer is $P=Q>I$.