Alternating Power Series - Limits

[ellynx]

1. Alternating power series question on convergence interval.

I'm wrestling a bit with an alternating power series, the teacher has the convergence interval to be
x = <-2,2] and I don't agree.

Homework Equations

Without further adue, here is the alternating power series in question:
\sum^{\infty}_{n=1}(-1)^{n-1}\frac{x^{n}}{n*2^{n}}
Ratio test says;
\lim_{n=\infty}|\frac{x^{n+1}}{(n+1)*2^{n+1}}\frac{n*2^{n}}{x^{n}}|
and end at the fruity limes below
\frac{|x|}{2}\lim_{n=\infty}|\frac{n}{(n+1)}|
now I understand that, \lim_{n=\infty}|\frac{n}{(n+1)}| \approx 1

and in order to satisfy the criteria, ratio < 1, the limits for x must lie in <-2,2>.

Inserting limits for x in the original series i got:

x=2: \sum^{\infty}_{n=1}(-1)^{n-1}\frac{1}{n}, converges
x=-2: \sum^{\infty}_{n=1}(-1)^{2n-1}\frac{1}{n}, diverges

The Attempt at a Solution

This far I agree with the limits stated by the teacher, but something bugs me, I computed the series with many different values for x>2 in maple and the result was a finite real number every time. For positive x values the series seems to alternate and converge, even though it brakes the ratio < 1. Is it possible that this series is conditionally convergent, or am I on the wrong mind track?

Cheers.

 

[estro]

HINT: Maybe it is worth to think about "Leibniz Criterion"?

 

[Dick]

HINT: Maybe it is worth to think about "Leibniz Criterion"?

In other words, the terms of (x/2)^n/n do not approach 0 as n->infinity if x>2. Show that if it will help. Sure they alternate, but that's not enough.

 

[estro]

Yes, but for other values root test decides right away. [almost trivial]

 

[Dick]

Yes, but for other values root test decides right away. [almost trivial]

Absolutely, so does the ratio test. I wasn't disagreeing with your answer in any way. I was just pointing out the part of the alternating series test that fails. If ellynx is picking values like x=2.01 the series will appear to decrease at first but if you get out past n=200 (I think), they will start increasing.

 

[ellynx]

Thank you for the replies and for shedding some light on this series business for me. I think I got it. I just have to pick the tests with care.

 

[Mark44]

Ratio test says;
\lim_{n=\infty}|\frac{x^{n+1}}{(n+1)*2^{n+1}}\frac{n*2^{n}}{x^{n}}|
and end at the fruity limes below
\frac{|x|}{2}\lim_{n=\infty}|\frac{n}{(n+1)}|

"end at the fruity limes...?"

now I understand that, \lim_{n=\infty}|\frac{n}{(n+1)}| \approx 1

No, the limit isn't approximately 1. It's true that for large n, n/(n + 1) is approximately 1, but in the limit, this ratio is exactly 1.

 

[ellynx]

Yes of course, it should say equal to one, thank you for correcting me Mark44.

as goes for "end at the fruity limes," its just a silly pun on "limes" as the fruit lime and of course as in limes or limit.

 

[Ray Vickson]

1. Alternating power series question on convergence interval.

I'm wrestling a bit with an alternating power series, the teacher has the convergence interval to be
x = &lt;-2,2] and I don't agree.

Homework Equations

Without further adue, here is the alternating power series in question:
\sum^{\infty}_{n=1}(-1)^{n-1}\frac{x^{n}}{n*2^{n}}
Ratio test says;
\lim_{n=\infty}|\frac{x^{n+1}}{(n+1)*2^{n+1}}\frac{n*2^{n}}{x^{n}}|
and end at the fruity limes below
\frac{|x|}{2}\lim_{n=\infty}|\frac{n}{(n+1)}|
now I understand that, \lim_{n=\infty}|\frac{n}{(n+1)}| \approx 1

and in order to satisfy the criteria, ratio < 1, the limits for x must lie in <-2,2>.

Inserting limits for x in the original series i got:

x=2: \sum^{\infty}_{n=1}(-1)^{n-1}\frac{1}{n}, converges
x=-2: \sum^{\infty}_{n=1}(-1)^{2n-1}\frac{1}{n}, diverges

The Attempt at a Solution

This far I agree with the limits stated by the teacher, but something bugs me, I computed the series with many different values for x>2 in maple and the result was a finite real number every time. For positive x values the series seems to alternate and converge, even though it brakes the ratio < 1. Is it possible that this series is conditionally convergent, or am I on the wrong mind track?

Cheers.

Maple computes the sum exactly as log(1 + x/2), which, of course, is true if |x| < 1. The series diverges for x > 2, but the *analytic continuation* of the series continues to make sense for x > 2. This type of thing is not at all uncommon: we start with a series that is valid definition of a function in a certain region, then can extend the function outside that region by changing the series.

For example, in your case we can expand f(x) = log(1 + x/2) about x = a to get
\displaystyle f(x) = log(1 + \frac{a}{2}) + \sum_{n=1}^{\infty} \frac{(-1)^{(n-1)}}{n} \frac{(x-a)^n}{(a+2)^n} If we set a = 1/2 we have two convergent series for f(x) near x = 1/2: the original one, and the new one. The new series converges for |x - 1/2| < 5/2, so converges for x out to 3.

RGV