f continuous on interval: show limit of sum is zero

BrainMan

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This problem appears on a practice final my professor gave out and I'm not sure what to do. Could someone show me how to do this limit?

Let f be continuous on the interval [0,1]. Show that:

lim as n approaches infinity of

(1/n) * [summation from k=1 to n of ((-1)^k)*[(f(k/n) - (f((k-1)/n)]]

equals zero.

I hope you can understand what I wrote. Thanks for any help you can offer.
 
BrainMan said:
(1/n) * [summation from k=1 to n of ((-1)^k)*[(f(k/n) - (f((k-1)/n)]]
The square brackets match (you have two open-brackets and two close-brackets), but you're missing two close-parentheses. Do you perhaps mean the following?

. . . . .\(\displaystyle \L \frac{1}{n}\, \sum_{k=1}^n \, \left[(-1)^k\, \left(f\left(\frac{k}{n}\right)\, -\, f\left(\frac{k-1}{n}\right)\right)\right]\)

Thank you! :D

Eliz.
 
Yes, that's exactly what I want.

Now how do you show the limit as n approaches infinity is zero for f continuous in [0,1]?
 
Fact #1. If f is continuous on [0,1] it is uniformly continuous there.

Fact #2. \(\displaystyle \left| {\sum\limits_{k = 1}^N {\left( { - 1} \right)^k \left[ {f\left( {\frac{k}{N}} \right) - f\left( {\frac{{k - 1}}{N}} \right)} \right]} } \right| \le \sum\limits_{k = 1}^N {\left| {\left[ {f\left( {\frac{k}{N}} \right) - f\left( {\frac{{k - 1}}{N}} \right)} \right]} \right|}\)

Fact #3. \(\displaystyle \left| {\frac{k}{N} - \frac{{k - 1}}{N}} \right| = \left| {\frac{1}{N}} \right|\).
 
How do I use fact #3? What you said makes sense; I just don't see how it all ties together.

Thanks a lot for replying before.
 
Do you know the definition of uniformly continuous?
\(\displaystyle \delta > 0\quad \Rightarrow \quad \left( {\exists N} \right)\left[ {\frac{1}{N} < \delta } \right]\).
A some value the division points are ‘close’ enough to use uniform continuity.
 
How do you know what the limit of the summation is?
I just don't see what you're getting at (keeping in mind, I could read about uniform continuity for three weeks (I've tried) and still not understand it).
 
Use uniform continuity.

\(\displaystyle \varepsilon > 0\quad \Rightarrow \quad \left( {\exists \delta > 0} \right)\left| {x - y} \right| < \delta \quad \Rightarrow \quad \left| {f(x) - f(y)} \right| < \varepsilon\).

\(\displaystyle \left( {\frac{1}{N}} \right)\left( {\sum\limits_{k = 1}^N {\left| {\left[ {f\left( {\frac{k}{N}} \right) - f\left( {\frac{{k - 1}}{N}} \right)} \right]} \right|} } \right) \le \left( {\frac{1}{N}} \right)\left[ {N\varepsilon } \right] = \varepsilon\)
 
Ok. I understand that perfectly, but I still don't see how it applies to the limit.

I really appreciate you helping me and being patient.
 
That shows that for any \(\displaystyle \varepsilon > 0\) the absolute value of the sum will become \(\displaystyle < \varepsilon\) which means the limit is 0.
 
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