How that a simple series diverges

[Mr Davis 97]

Homework Statement

Show that the series $\displaystyle \sum_0^{\infty} (-1)^nn$ diverges

Homework Equations

The Attempt at a Solution

It seems obvious that it diverges, for although the terms oscillate, they get bigger and bigger and never really cancel each other out. However, I am not sure how to show rigorously that it diverges, that is, using one of the tests from calc 2. I looked at a possible solution, and it said, using the Test for Divergence, we could conclude that since $\displaystyle \lim_{n \to \infty} |(-1)^n n| = \infty$, the series diverges. But are we justified in using the absolute value? I thought that the test for divergence would just be $\displaystyle \lim_{n \to \infty} (-1)^n n$, which doesn't seem to be able to be evaluated.

 

[andrewkirk]

Use the triangle inequality to prove that, if the sum converges, the distance between successive partial sums must tend to zero. You can get that from the fact that the distance form both partial sums to any putative limit must tend to zero.

 

[scottdave]

In order for the sum to converge, one of the criteria is that the sequence of terms must converge to zero. I believe how that is how it goes. I am on my phone and don't have a book handy.

 

[Mark44]

Homework Statement

Show that the series $\displaystyle \sum_0^{\infty} (-1)^nn$ diverges

Homework Equations

The Attempt at a Solution

It seems obvious that it diverges, for although the terms oscillate, they get bigger and bigger and never really cancel each other out. However, I am not sure how to show rigorously that it diverges, that is, using one of the tests from calc 2. I looked at a possible solution, and it said, using the Test for Divergence, we could conclude that since $\displaystyle \lim_{n \to \infty} |(-1)^n n| = \infty$, the series diverges. But are we justified in using the absolute value? I thought that the test for divergence would just be $\displaystyle \lim_{n \to \infty} (-1)^n n$, which doesn't seem to be able to be evaluated.

Just use the Test for Divergence. It says that if $\lim a_m \ne 0$ or the limit fails to exist, the series diverges. What is $\lim_{n \to \infty} (-1)^n n$?

In order for the sum to converge, one of the criteria is that the sequence of terms must converge to zero. I believe how that is how it goes.

That's a necessary criterion, but not a sufficient one. The obvious example is the harmonic series $\sum_{n = 1}^\infty \frac 1 n$.

 

[scottdave]

Since it failed the Divergence test, you have proved that the sum diverges. No need for more tests (like with the Harmonic Series)

 

[Mark44]

Since it failed the Divergence test, you have proved that the sum diverges.

No, the OP showed that $\sum_{n = 0}^\infty |(-1)^n n|$ diverged, but that wasn't the series he was given, which was $\sum_{n= 0}^\infty (-1)^n n$.

For example, it can be shown that $\sum_{n = 1}^\infty |\frac{(-1)^n}n|$ diverges, but $\sum_{n = 1}^\infty \frac{(-1)^n}n$ converges.

No need for more tests (like with the Harmonic Series)

I gave that as an example that $\lim a_n = 0$ was necessary, but not sufficient. I wasn't saying that the OP needed to look at additional tests.

 

[scottdave]

It seemed to me that the OP, @Mr Davis 97 was looking for more tests, giving the Harmonic Series as an example of where the sequence converges to zero, but the series diverges. Since he already knows that the sequence does not converge, that shows that the series diverges.

 

[Mark44]

It seemed to me that the OP, @Mr Davis 97 was looking for more tests, giving the Harmonic Series as an example of where the sequence converges to zero, but the series diverges. Since he already knows that the sequence does not converge, that shows that the series diverges.

I think that we have different interpretations of what the OP said.

It seems obvious that it diverges, for although the terms oscillate, they get bigger and bigger and never really cancel each other out. However, I am not sure how to show rigorously that it diverges, that is, using one of the tests from calc 2. I looked at a possible solution, and it said, using the Test for Divergence, we could conclude that since $\displaystyle \lim_{n \to \infty} |(-1)^n n| = \infty$, the series diverges. But are we justified in using the absolute value? I thought that the test for divergence would just be $\displaystyle \lim_{n \to \infty} (-1)^n n$, which doesn't seem to be able to be evaluated.

He intuitively believed that his series diverged, but didn't know how to show it. He used the Nth Term Test on a different series (i.e., $\sum_{n = 1}^\infty |(-1)^n n|$), but my point was that in this case, what the series of absolute values did wasn't relevant to the series he was given.

He said that he didn't think that $\lim_{n \to \infty} (-1)^n n$ could be evaluated, but since this limit doesn't exist, the Nth Term Test shows that the series diverges.

Edit: Removed erroneous text that Alt. Series Test could also be used to show divergence.

 

[vela]

Since his original series is an alternating series, the Alternating Series Test could be also used to show the same result.

Did you mean to say this? The alternating series test can prove that a series converges, but I don't recall it being able to prove that the series diverges.

 

[Mark44]

Did you mean to say this? The alternating series test can prove that a series converges, but I don't recall it being able to prove that the series diverges.

You're right. I misremembered the fine print of the Alt. Series Test, and was thinking that if the conditions weren't met, the series diverged. I have revised my earlier post.