# Homework Help: Wierstrauss m test caucy criteria

1. Nov 29, 2008

1. The problem statement, all variables and given/known data
So, I missed class the day Weierstrass M-test and cauchy criterion. I have a theorem for Weierstrass M-test, but the book gives no examples on how to use it. I figure cauchy criterion has to do with series, and states that a series is convergent if the partial sums is cauchy. If this one is wrong please let me know. I made it up, just now. We should be in the sequence and series of functions chapter, and there are is a lot of things that look cauchy (fn(x) - f(x)) < epsilon, but none called "cauchy criterion for series" like the email I get.

2. Relevant equations
Suppose {fn} is a sequence of functions defined on E, and {Mn} is a sequenc of nonnegative real numbers such that | fn(x) | <= Mn for all x in E, all positive integers n. If
the Sum Mn converges then the sum of fn converges uniformly.

3. The attempt at a solution
My guess is that how to use Weierstrass M test is sort of like a comparison test to derive uniform convergence. If you can get a series like 1/x^2 to converge and the sequence of functions is always less than 1/x^2, then the sequence of functions is uniformly convergent. Is this sounding right? There is no homework on this, so I don't have any examples to solve.

I'm really confused, they did a chapter in a day, which means I probably am not expected to get the entire depth down, but I get sad if I can't get it all the way.

2. Nov 29, 2008

### Office_Shredder

Staff Emeritus
You have the Cauchy criterion correct, but it's often written as the series converges iff for all epsilon > 0, there exists N such that n,m>N implies

$$| \sum_{k=m}^{n}a_k | < \epsilon$$

(which is equivalent of course, but just written out in a more conventional form)

The M-test you have the right idea... for example, I take the sequence of functions fn = 1/xn for 1/2<x<3/4

Then

$$\sum_{n=1}^{\infty}f_n(x)$$ converges uniformly to $$\frac{1}{1-x}$$

Proof:

Clearly it converges pointwise (geometric series). It's uniform as for any x, |fn(x)|<= (3/4)n which we'll call Mn

Clearly $$\sum_{n=1}^{\infty}M_n$$ converges, and hence the series
$$\sum_{n=1}^{\infty}f_n(x)$$ converges uniformly.

Notice the domain is important here... if we had picked 0<=x<1, this would have failed as we would have gotten the smallest possible value for each Mn would be 1, and $$\sum_{n=1}{\infty}M_n$$ no longer converges, and you can't conclude the series converges uniformly to 1/(1-x)

You may have noticed that conveniently our point picked was one of the endpoints of the domain we were looking at.... this is often the case, since the M-test is usually applied to power series, and each monomial term is going to be monotonically increasing.

Wikipedia has another example

http://en.wikipedia.org/wiki/Uniform_convergence#Exponential_Function

it's done over the complex numbers, but if instead of open disks you substitute open intervals, and instead of C you substitute R, it's exactly the same

3. Nov 30, 2008