ISI2015-MMA-81

Question

If $f$ is continuous in $[0,1]$ then $$\displaystyle \lim_ {n \to \infty}   \sum_{j=0}^{[n/2]} \frac{1}{n} f \left(\frac{j}{n} \right)$$ (where $[y]$ is the largest integer less than or equal to $y$)

• A. does not exist
• B. exists and is equal to $\frac{1}{2} \int_0^1 f(x) dx$
• C. exists and is equal to $ \int_0^1 f(x) dx$
• D. exists and is equal to $\int_0^{1/2} f(x) dx$

Answer 1

Let $f$ be continuous on $[0,1]$.

$$\lim_{n\rightarrow \infty} \sum_{j=0}^{[n/2]} \frac{1}{n} f\left(\frac{j}{n}\right)$$
where [n] is greatest integer function.

I used the following relation,

$$\int_{a}^{b}f(x)\ dx =\lim_{\delta_x \rightarrow 0} \sum_{x=a}^{b}f(x)\delta_x$$

I considered $x=\frac{j}{n}$ and $\delta_x = \frac{(j+1)-j}{n}=\frac{1}{n}$.

At $j=0$, I've $x=0$.

At $j=[n/2]$, if $n$ is even then $x=\frac{1}{2}$. If $n$ is odd,$[n/2]=\frac{n-1}{2}$ then $x=\frac{1}{2}[1-\frac{1}{n}]$.

Therefore, as $n\rightarrow \infty$ out $x=\frac{1}{2}$.

$$\lim_{n\rightarrow \infty} \sum_{j=0}^{[n/2]} \frac{1}{n} f\left(\frac{j}{n}\right)=\int_{0}^{1/2} f(x)\ dx$$