Limit of a continuous function with a summation

[Ozma]

Let [imath]f:[0,1]\to\mathbb{R}[/imath] be a continuous function. Evaluate [imath]\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n (-1)^k f\left(\frac{k}{n}\right)[/imath].

I tried to estimate [imath]1 \le k \le n \implies 0<\frac{1}{n} \le \frac{k}{n} \le 1[/imath] and so, by the continuity hypothesis, [imath]f[/imath] is continuous at [imath]k/n[/imath] for any [imath]k=1,2,\dots,n[/imath] and for any [imath]n\in\mathbb{N}[/imath], however this hasn't helped me to estimate the summation with the continuity in a useful way. Can I have a hint?

 

[Steven G]

You have (1/n)(-f(1/n)) + f(2/n) - f(3/n) + ... +/- f(1)) = [f(2/n)-f(1/n)]/(1/n) + [f(4/n)-f(3/n)]/(1/n) + ... + [f(n/n)-f(1-1/n)]/(1/n).

The above may not always be correct. Why not? Does this help?

 

[blamocur]

What happens if you add [imath]\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^n f\left(\frac{k}{n}\right)[/imath] ?

 

[Dr.Peterson]

Let [imath]f:[0,1]\to\mathbb{R}[/imath] be a continuous function. Evaluate [imath]\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n (-1)^k f\left(\frac{k}{n}\right)[/imath].

I tried to estimate [imath]1 \le k \le n \implies 0<\frac{1}{n} \le \frac{k}{n} \le 1[/imath] and so, by the continuity hypothesis, [imath]f[/imath] is continuous at [imath]k/n[/imath] for any [imath]k=1,2,\dots,n[/imath] and for any [imath]n\in\mathbb{N}[/imath], however this hasn't helped me to estimate the summation with the continuity in a useful way. Can I have a hint?

In addition to the hints you've been given, you might experiment by picking a simple function for f, say f(x)=x, or f(x)=x^2, and working out the sum, to see if these suggest anything.

And you might graph the function and what area the summation represents (particularly using Steven's hint).

 

[blamocur]

You have (1/n)(-f(1/n)) + f(2/n) - f(3/n) + ... +/- f(1)) = [f(2/n)-f(1/n)]/(1/n) + [f(4/n)-f(3/n)]/(1/n) + ... + [f(n/n)-f(1-1/n)]/(1/n).

The above may not always be correct. Why not? Does this help?

Am I missing something or have you switched from multiplication by [imath]\frac{1}{n}[/imath] to division?

 

[Steven G]

Am I missing something or have you switched from multiplication by [imath]\frac{1}{n}[/imath] to division?

Ooops