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## Homework Statement

Which of the following series is point-wise convergent, absolutely convergent? Which ones are ##L^2(-\pi,\pi)##-convergent.

A) ##\sum_1^\infty \frac{\cos n \theta}{n+1}##

B) ##\sum_1^\infty \frac{(-1)^n\cos n \theta}{n+1}##

## Homework Equations

Abel's test:[/B]

Suppose ##\sum a_n## is a convergent series , ##\{ b_n \}## is a monotone sequence and bounded then ##\sum a_nb_n## is also convergent.

**Dirichlet's test:**

If ##\{a_n\}## is a sequence of real numbers and ##\{ b_n \}## a sequence of complex numbers satisfying

##a_n \ge a_{n+1}##

##\lim_{n\to \infty} a_n = 0##

##\left| \sum_1^N b_n \right| \le M## where ##M## is some constant then

##\sum_1^\infty a_nb_n## converges.

## The Attempt at a Solution

To prove A is convergent by applying the Dirichlet test I need to show that every partial sum of ##\cos n \theta## is bounded. (excluding ##\theta=0## since then the series obviously diverges.) Writing ##\cos n\theta= \frac{e^{in\theta}+ e^{-in\theta}}{2}## we have two geometric series that are bounded, hence ##\sum \cos n \theta## is bounded.

Since ##\frac{1}{n+1}## is decreasing and and approaches ##0## the series A is convergent. I suspect the series ain't absolutely convergent but I'm not sure how to prove that, If the series was convergent I guess I could use weierstrass M test too prove that but I suspect it's divergent.

The next series (B) is convergent since ##\sum \frac{(-1)^n}{1+n}## is convergent by the alternating series test and it then follows from Abel's test that the series converge since ##\sum \cos n\theta## is bounded (for ##\theta \ne 0##). Is the above correct? For absolute convergence I got the same problem as for (A) which I don't know how to prove.

For the part about ##L^2##-convergence I'm not sure what is meant by the question. It isn't supposed to be ##l^2## convergence? Or do I have a integral around that like ##\int \sum |f_n(\theta)| < \infty##?