TITLE: find the value of L?
QUESTION [5 upvotes]: Suppose  f is continuous on $[0,1]$. Find 
$$L=\lim_{n\rightarrow \infty} \frac{1}{n}  \sum_{k=1}^{n} (-1)^k f(\frac{k}{n}) $$
My attempts :   I was thinking  about Riemann sum... but  $(-1)^k$  creates  confusion.
Thanks in advance.

REPLY [2 votes]: If $f$ is continuous on $[0,1]$ it is uniformly continuous; for any $n\geq 1$ let $$\Delta_n = \sup_{\substack{x,y\in[0,1]\\|x-y|\leq \frac{1}{n}}}\left|f(x)-f(y)\right|$$
and $M=\sup_{x\in[0,1]}\left|f(x)\right|$. By the triangle inequality
$$ \left|\sum_{k=1}^{n}(-1)^k f\left(\tfrac{k}{n}\right)\right| \leq M + \frac{n}{2}\Delta_n $$
and since $\Delta_n \to 0$ as $n\to +\infty$, the wanted limit is zero.