Weierstrass M Test: Finding the Radius of Convergence

[latentcorpse]

I just want to check something here.

If we want the radius of convergence of \sum_{n=0}^{\infty} x^n, we cannot use the ratio test because its not a series, it's a series of functions i.e. we have \sum f_n(x) not \sum a_n. is this true?

so if we apply the M test to the f_n, then we need to find a sequence of positive integers M_n such that |f_n(x)| \leq M_n \forall x \in E where E is our interval.
Then we notice that such an M_n cannot exist for |x|>1 and hence for the given power series, teh radius of convergence is R=1.

how does this look to everyone?

 

[e(ho0n3]

If we want the radius of convergence of \sum_{n=0}^{\infty} x^n, we cannot use the ratio test because its not a series, it's a series of functions i.e. we have \sum f_n(x) not \sum a_n. is this true?

Depends on whether x is fixed or not. If it is, then it's a series (of numbers). If it isn't, then it's a series of functions. But you want the radius of convergence and the ratio and root tests are the only tools I know of to calculate that so start by fixing x.

so if we apply the M test to the f_n, then we need to find a sequence of positive integers M_n such that |f_n(x)| \leq M_n \forall x \in E where E is our interval.
Then we notice that such an M_n cannot exist for |x|>1 and hence for the given power series, teh radius of convergence is R=1.

The conclusion of the M-test has nothing to do with radius of convergence. Try again.

 

[latentcorpse]

this is trivial by the ratio test |\frac{x^{n+1}}{x^n}|=|x| which converges if |x|<1 so the radius of convergence is 1 - actually would this not also converge for x=1 by examination, so it converges on (-1,1]

but how do we know x is to be fixed? it doesn't say anything about it in the question so it's probably ok to assume this like you've done but we do have a whole chapter dedicated to sequences of functions.

 

[e(ho0n3]

this is trivial by the ratio test |\frac{x^{n+1}}{x^n}|=|x| which converges if |x|<1 so the radius of convergence is 1 - actually would this not also converge for x=1 by examination, so it converges on (-1,1]

You already know that it converges if |x| < 1. Why bother worrying if x = 1? In any case, if x = 1, the series is 1 + 1 + ... which clearly diverges.

but how do we know x is to be fixed?

Because you want to know for what values of x the series converges. Analogously, when you solve ax + b = 0 for x, you assume that x is some fixed value and then manipulate the equation to determine what it is.

 

[Billy Bob]

(1) Pointwise remark: Convergence, absolute convergence, or divergence of \sum f_n(z) or \sum a_n z^n is always a pointwise concern. Regard z as a fixed number, and use comparison test, ratio test, root test, or any test which works.

(2) "Function of z" remark: If g(z)=\sum f_n(z) or h(z)=\sum a_n z^n is to be used as a function of z, e.g. differentiate it, integrate it, or decide if it is a continuous function, then you need uniform convergence.

The M test is a "beefed up" comparison test can be used on g(z) or h(z) to determine not only absolute convergence, but uniform convergence.

The ratio test has its own "beefed up" version, but it is still called the ratio test, because it is still the same test. It can be used on h(z) but not on g(z). For h(z), the series converges absolutely for all z in the open disk of convergence, and furthermore it converges uniformly on any closed disk inside the open disk of convergence. The ratio test can be used to determine R.

The "elementary" ratio test (that doesn't mention uniform convergence) can be proved by using the comparison test, comparing to a geometric series.

The "advanced" ratio test (that includes uniform convergence) can be proved simply by looking at the proof of the "elementary" version and observing that when you use the comparison test, you are really using the M test.

Moral: the difference between M test and ("advanced") ratio test is that M test can be used on g(z) or h(z), but ratio test only on h(z).

Note: of course when I say h(z)=\sum a_n z^n, I also should include h(z)=\sum a_n (z-c)^n.