Exercise 6.5.2 in our text is one of those example type problems that make great quiz questions. For each part, write down the statement here and provide an example.

# The examples of exercise 6.5.2

Just to get the ball rolling... Exercise 6.5.2 states:

Find suitable coefficients $(a_{n})$ such that the resulting power series $\sum a_{n} x^{n}$

(a) converges absolutely for all $x \in [-1,1]$ and diverges off this set

(b) converges conditionally at $x=-1$ and diverges at $x=1$

(c) converges conditionally at both $x=-1$ and $x=1$

(d) Is it possible to find an example of a power series that converges conditionally at $x=-1$ and converges absolutely at $x=1$.

I'm going to go ahead and say that (d) is a no. My reasons are as follows...

Absolute convergence is: If $\sum a_{n}$ and $\sum \vert a_{n} \vert$ both converge, than $\sum \vert a_{n} \vert$ is absolutely convergent.

Since $ \vert -1 \vert = 1$, then the power series will absolutely converge at both $x=1$ and $x=-1$.

I don't know if this is right and I'm having a hard time explaining myself. If anyone wants to correct/critique me please do!

@ediazloa Thanks for getting the ball rolling!

As for your thoughts on part (d) - First, I agree with your *answer*. Second, I like your *intuition*. But, I still think we're a bit short of a definitive explanation. When confronted with this type of question, I recommend thumbing through the text to find (hopefully) relevant theorems - perhaps around the, oh, let's say page 170ish.

Regarding (a), I think an answer that we brought up early on (outside of Discourse) is correct. I only now really understand why.

The series

$$\sum_{n=1}^{\infty} \frac{x^n}{n^2}$$

should be sufficient. It converges absolutely on $[-1,1]$ due to the Weierstrauss $M$ test ($x^n/n^2 \leq 1/n^2$ for $x$ on this interval, and $\sum \frac{1}{n^2}$ converges). It is also divergent away from this set, as $x^n/n^2$ does not go to zero as $n \rightarrow \infty$.

Adding to what you have @ediazloa and using @mark 's suggestion:

If the power series converges absolutely at $x = 1$, using Thm. 6.5.2, we can say that it converges *uniformly* on the closed interval $[-1,1]$. Therefore it would not converge conditionally at $x = -1$.

@cseagrav In fact, that was exactly what I was thinking. As I think about it more closely, however, I realize that I read that theorem too fast. Actually, it is possible that a power series converges uniformly on a closed interval but only conditionally at one endpoint. For example,

$$\sum\frac{x^n}{n}$$

converges uniformly by Abel's theorem on $[-1,0]$ but only conditionally at $x=-1$.

As I think about it more, I think that @ediazloa was *very* close. In fact, $|-x|=|x|$ so, if a series converges absolutely at $x$, then it must converge absolutely at $-x$!

@mark It did seem as though her explanation was sufficient and I see now the error of using Thm. 6.5.2 a bit too quickly through the counter example. To make sure I understand, since we can say $|1| = |-1|$ and it converges absolutely at $1$, then it converges absolutely at $-1$.

Your counter example

$$\sum_{n=1}^{\infty} \frac{x^n}{n}$$

is a good answer for part b) also, since it converges conditionally at $x = -1$ but diverges at $1$.

I am just not sure what justification is needed here, if any.

@cseagrav I still think it's a little more thant $|1|=|-1|$. You've got a series

$$\sum a_nx^n.$$

It converges absolutely at $1$ if

$$\sum |a_n 1^n| = \sum |a_n|$$

converges and it converges absolutely at $-1$ if

$$\sum |a_n (-1)^n| =\sum |a_n|\times|(-1)^n| =\sum |a_n|$$

converges. Thus, you're ultimately checking convergenge of the same series.