Divergence of a particular serie

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Homework Statement


\sum_{n=1}^{\infty}{(-1)^n*n}

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 b_{n+1} > b_n, and not the opposite.

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

What am I doing wrong?

Thanks in advance.
 
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ShizukaSm said:

Homework Statement


\sum_{n=1}^{\infty}{(-1)^n*n}

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.
ShizukaSm said:
, what I tried to do so far:

Alternating testing series: Of course, my first thought, the test fails because b_{n+1} > b_n, and not the opposite.
This test doesn't fail - it is not applicable here.
ShizukaSm said:
Then I tried the test for divergence:
\lim_{n->\infty}{(-1)^n*n}
But I can't solve this limit, I could solve it without the (-1)^n. 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.
ShizukaSm said:
What am I doing wrong?

Thanks in advance.
 
Last edited:
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 said:
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.
ArcanaNoir said:
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.
 
ShizukaSm said:
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".
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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