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## SEEMOUS 2011 Problem 4

Let $f :[0,1]\to \Bbb{R}$ be a twice continuously differentiable increasing function. Define the sequences given by
$L_n=\displaystyle \frac{1}{n}\sum_{k=0}^{n-1}f\left(\frac{k}{n}\right)$ and
$U_n=\displaystyle \frac{1}{n}\sum_{k=1}^{n}f\left(\frac{k}{n}\right)$ for $n \geq 1$.
The interval $[L_n,U_n]$ is divided into three equal segments. Prove that for large enough $n$, the number $I=\displaystyle \int_0^1 f(x)dx$ belongs to the middle one of these segments.
SEEMOUS 2011 Problem 4

Solution: Define $F(x)=\displaystyle \int_0^x f(t)dt$. Then we can write $I=F(1)-F(0)=F(1)-F(\frac{n-1}{n})+F(\frac{n-1}{n})-F(\frac{n-2}{n})+...+F(1/n)-F(0)$. We apply Taylor’s expansion formula of order 2 for $F$ in points $0,1/n,...,\frac{n-1}{n}$.

We have $F(x)=F(k/n)+(x-k/n)f(k/n)+\frac{(x-k/n)^2}{2}f^\prime(\theta),\ \theta \in [k/n,x],\ k=0...n-1$. For $x=1/n,...,1$ we get $F(\frac{k+1}{n})-F(\frac{k}{n})=\frac{1}{n}f(\frac{k}{n})+\frac{1}{2n^2} f^\prime(\theta_k),\ \theta_k \in [\frac{k}{n},\frac{k+1}{n}]$. Summing this for $k=0,1,...,n-1$ we get $I=L_n+\frac{1}{2n}\sigma_n$, where $\sigma_n$ is a Riemann sum for $\int_0^1 f^\prime(t)dt=f(1)-f(0)$.

The middle intervals are $[\frac{2}{3}L_n+\frac{1}{3}U_n,\frac{1}{3}L_n+\frac{2}{3}U_n]=[u_n,v_n]$. If $f(1)=f(0)$ the function is constant, and we know that $f$ is increasing and therefore, non-constant (increasing means strictly increasing). Therefore $f(1)>f(0)$.

Now, we calculate $n(I-u_n)= \frac{1}{3}(L_n-U_n)+\frac{1}{2}\sigma_n=\frac{1}{3}(f(0)-f(1))+\frac{1}{2}\sigma_n$, and taking the limit for $n \to \infty$ we get that $n(I-u_n)\to \frac{1}{6}(f(1)-f(0))>0$, which means that for $n$ great enough we have $I>u_n$. On the other hand, $n(v_n-I)=\frac{2}{3}(U_n-L_n)-\frac{1}{2}\sigma_n\to\frac{1}{6}(f(1)-f(0))>0$.