Prove $\int_{0}^{n}\dfrac{ax}{a+x}dx<\sum_{k=1}^n \dfrac{ak}{a+k}<\int_{0}^{n+1}\dfrac{ax}{a+x}dx$ Let n be a positive integer.
When $a$ is a positive integer between $1$ and $n$.
Prove
$$\int_{0}^{n}\dfrac{ax}{a+x}dx<\sum_{k=1}^n \dfrac{ak}{a+k}<\int_{0}^{n+1}\dfrac{ax}{a+x}dx$$
I've spent the last few hours showing this by using the inequality
$\dfrac{ak}{a+k}\leq\dfrac{ax}{a+x}\leq\dfrac{a(k+1)}{a+k+1}$ and integrating it.
But I failed.
 A: Define
$$f:[0,n]\to\mathbb{R},\quad x\mapsto\frac{ax}{a+x},$$
and notice that $f$ is a strictly increasing function. Now consider $g$ defined by
$$g(x)=f(\lceil x\rceil).$$
As $f$ is increasing, this results in the inequality $f(x)\leq g(x)$, and it is easy to find intervals where the inequality is strict, which yields that
$$\int_0^n f(x)~\mathrm{d}x<\int_0^n g(x)~\mathrm{d}x.$$
But now notice that, since we are dealing with the ceiling function, if $k\in\mathbb{N}$, and $x\in(k-1,k]$, then $g(x)=f(k)$, and so we can write
$$\int_0^n g(x)~\mathrm{d}x=\sum_{k=1}^n\int_{k-1}^kg(x)~\mathrm{d}x=\sum_{k=1}^n\int_{k-1}^k f(k)~\mathrm{d}x=\sum_{k=1}^nf(k).$$
This establishes the first inequality,
$$\int_0^n f(x)~\mathrm{d}x<\sum_{k=1}^nf(k).$$
I'll leave the second inequality to you, for which you should be able to do a very similar argument.
A: Some hints.
$$\frac{ax}{a+x}=a-\frac{a^2}{a+x}$$Is a strictly increasing function in $x$. By the linearity of the integral: $$\int_0^n=\sum_{k=1}^n\int_{k-1}^k$$On each strip $[k-1,k]$ you can make some bounds, using the “strictly increasing” part.
You can conceptualise this as drawing the graph of the function and covering its graph by fat, unit length, rectangles, bounding (depending on the convexity/concavity of the function) above or below. For a better visual depiction, look up the integral test for divergent/convergent series.
