An archived instance of discourse for discussion in undergraduate Real and Numerical Analysis.

# The examples of exercise 6.5.2

mark

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.

ediazloa

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!

mark

@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.

Cromer

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 $n \rightarrow \infty$.

cseagrav

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 $x = -1$.

mark

@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 $-x$!

cseagrav

@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.

mark

@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.