Divergence of a particular serie

[ShizukaSm]

Homework Statement

[itex]\sum_{n=1}^{\infty}{(-1)^n*n}[/itex]

The Attempt at a Solution

Well, this is a series I came upon when analyzing the endpoints of a particular power series, the thing is, my book says it's divergent by the Test for divergence, however, I can't find this result, what I tried to do so far:

Alternating testing series: Of course, my first thought, the test fails because [itex]b_{n+1} > b_n[/itex], and not the opposite.

Then I tried the test for divergence:
[itex]\lim_{n->\infty}{(-1)^n*n}[/itex]
But I can't solve this limit, I could solve it without the [itex](-1)^n[/itex]. But I can't just "take out the term I don't like and solve it".

What am I doing wrong?

Thanks in advance.

 

[Mark44]

Homework Statement

[itex]\sum_{n=1}^{\infty}{(-1)^n*n}[/itex]

The Attempt at a Solution

Well, this is a series I came upon when analyzing the endpoints of a particular power series, the thing is, my book says it's divergent by the Test for divergence, however, I can't find this result

Some books call this the "Nth term test for divergence." This theorem is very simple to use, and should probably be used first. It says that in a series ##\sum a_n##, if ##\lim_{n \to \infty} a_n ≠ 0##, then the series diverges.

Note that this theorem can be used only on series for which ##\lim_{n \to \infty} a_n ≠ 0##. If the limit equals zero, the series could converge or it could diverge.

, what I tried to do so far:

Alternating testing series: Of course, my first thought, the test fails because [itex]b_{n+1} > b_n[/itex], and not the opposite.

This test doesn't fail - it is not applicable here.

Then I tried the test for divergence:
[itex]\lim_{n->\infty}{(-1)^n*n}[/itex]
But I can't solve this limit, I could solve it without the [itex](-1)^n[/itex]. But I can't just "take out the term I don't like and solve it".

Expand the series by writing a few of the terms. It should be obvious what this series is doing.

What am I doing wrong?

Thanks in advance.

 

[ShizukaSm]

Yes indeed, I can see that it obviously diverge if I expand the series, I even tried graphing a few terms to look at that, but what I meant was, is there a "formal way" to write that?

 

[ArcanaNoir]

I think the divergence test says "if the limit is not 0 or if the limit does not exist, then the series diverges."

Here's a website to back that up: http://www.mathscoop.com/calculus/infinite-sequences-and-series/nth-term-test.php
I'd check a calculus book but I don't have one with me.

 

[Mark44]

I think the divergence test says "if the limit is not 0 or if the limit does not exist, then the series diverges."

If the limit does not exist, it is automatically not zero, so it isn't necessary to include the part about it not existing. Perhaps it is added above for clarity.

Here's a website to back that up: http://www.mathscoop.com/calculus/infinite-sequences-and-series/nth-term-test.php
I'd check a calculus book but I don't have one with me.

 

[ArcanaNoir]

Yes indeed, I can see that it obviously diverge if I expand the series, I even tried graphing a few terms to look at that, but what I meant was, is there a "formal way" to write that?

You would write something like "since the limit doesn't exist, the series diverges by the divergence test".