Series vs Series

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nasi112

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While I was analyzing and solving some infinite series, I came across these two series.

[math]\sum_{n=1}^{\infty}\frac{1}{n}[/math]
[math]\sum_{n=1}^{\infty}(-1)^{n}\frac{1}{n}[/math]
The first series is known to diverge but the second series counterintuitively converges. You as experts may say there's no counterintuitiveness behind the convergence, just apply the alternating series test.

[math]\frac{1}{n + 1} \leq \frac{1}{n}[/math]
[math]\frac{n}{n + 1} \leq 1[/math]
It's clear to me this inequality is true for all [imath]n \geq 1[/imath]. Today I accidently applied the limit comparison test on the alternating series and found that it diverges. Why the alternating series test is stronger than the limit comparison test in this situation? I mean by this why the limit comparison test doesn't count.
 
nasi112 said:
Today I accidentally applied the limit comparison test on the alternating series and found that it diverges. Why the alternating series test is stronger than the limit comparison test in this situation? I mean by this why the limit comparison test doesn't count.
Click to expand...
Presumably you applied the test incorrectly. If a test actually showed that it diverges, then it can't converge. That is different than a test not showing that it does converge. These are two very different ways for a test to "fail". You may be missing the details of what each test shows, or of when it can be applied (which is an important reason to know the actual theorems, and not just the general ideas).

We need to see what you did in order to point out with certainty what you did wrong.

Here are the two comparison tests:

Calculus II - Comparison Test/Limit Comparison Test

In this section we will discuss using the Comparison Test and Limit Comparison Tests to determine if an infinite series converges or diverges. In order to use either test the terms of the infinite series must be positive. Proofs for both tests are also given.
tutorial.math.lamar.edu

1750443643233.png
1750443572937.png
Note that the green parts must be true in order to apply the tests, and then the yellow parts give the conclusion.
 
This is how I used the test. A way I learned from blackpenredpen.

[math]\lim_{n\to \infty}\frac{1}{n}\frac{1}{\frac{1}{n}} = 1 > 0[/math], then the series diverges.
 
nasi112 said:
This is how I used the test. A way I learned from blackpenredpen.

[math]\lim_{n\to \infty}\frac{1}{n}\frac{1}{\frac{1}{n}} = 1 > 0[/math], then the series diverges.
Click to expand...
Please say in words what you are doing there, and why. There are several things wrong with it!

I assume you are trying to show that [imath]\sum_{n=1}^{\infty}(-1)^{n}\frac{1}{n}[/imath] diverges, right?

And did you compare what you are doing with what the theorem I quoted says?
 
I'll include [imath]a_n[/imath] and [imath]b_n[/imath] in the calculation to illustrate my process without using words.

[math]\sum_{n=1}^{\infty}(-1)^{n}\frac{1}{n} = \sum_{n=1}^{\infty}(-1)^{n}a_n[/math]
[math]\sum_{n=1}^{\infty}\frac{1}{n} = \sum_{n=1}^{\infty}b_n[/math]
[math]\lim_{n\to \infty}\frac{a_n}{b_n} = \lim_{n\to \infty} \frac{\frac{1}{n}}{\frac{1}{n}} = 1 > 0[/math]
[math]a_n, b_n > 0[/math] for all [imath]n[/imath]
 
Why are you applying the limit comparison test to [imath]a_n[/imath] (which is not the actual element of the series) and not to [imath](-1)^n a_n[/imath] ?
 
nasi112 said:
I'll include [imath]a_n[/imath] and [imath]b_n[/imath] in the calculation to illustrate my process without using words.

[math]\sum_{n=1}^{\infty}(-1)^{n}\frac{1}{n} =\sum_{n=1}^{\infty}{\color{Red}(-1)^{n}a_n}[/math]
[math]\sum_{n=1}^{\infty}\frac{1}{n} = \sum_{n=1}^{\infty}b_n[/math]
Click to expand...
Compare that to the theorem:

1750509736536.png

Did you apply the theorem as stated?

Can the theorem be applied to this series?
 
I'm confused. You guys mean I can't apply the limit comparison test to the alternating sum or I must apply it to whole series as [imath]a_n = (-1)^n\frac{1}{n}[/imath]?
 
nasi112 said:
I'm confused. You guys mean I can't apply the limit comparison test to the alternating sum or I must apply it to whole series as [imath]a_n = (-1)^n\frac{1}{n}[/imath]?
Click to expand...

If you analyze the proofs of the various tests for series to converge, then it always comes down to the comparison test. Different tests only save a couple of steps you would have to do if you wanted to go down to the comparison test. The problem with alternating series is, that order matters! Riemann's series theorem states that given any number [imath] c, [/imath] then there exists a rearrangement of the alternating series so that it converges to [imath] c. [/imath]

I think this theorem is the answer to your original question in post #1, if I correctly understood it. You can find an example in this article about series.

How you made the alternating series divergent remains a mystery to me. Can you show (all of) your calculations?
 
nasi112 said:
I'm confused. You guys mean I can't apply the limit comparison test to the alternating sum or I must apply it to whole series as [imath]a_n = (-1)^n\frac{1}{n}[/imath]?
Click to expand...
Both.

The theorem itself says it doesn't apply if the terms are not all positive. So, no, you can't apply it to any alternating series.

If you try to apply it, then you have to define [imath]a_n[/imath] as it is defined in the theorem, namely as the entire term. When you do so, you see that it immediately fails the first condition, that the terms be positive.

Instead, you apply the alternating series test, as you already know.

This is why I have emphasized that you need to learn the actual theorems, not just their general ideas.

You also need to understand how theorems work. If a theorem doesn't apply, then you just set it aside and look for one you can use. (Or sometimes you can make an adaptation -- for example, if not all the terms of a series are positive, but they are after some term, then you may apply the theorem to the latter series.)
 
Thanks Doctor. Got it.

Reply to the fresh_42
I applied the limit comparison test, and you can find my calculation in the 6th post. I determined that the limit is 1 > 0, which suggests that the series diverges because 1/n diverges. However, the doctor clarified why this test doesn't apply to the alternating sum.
 
nasi112 said:
Reply to the fresh_42
I applied the limit comparison test, and you can find my calculation in the 6th post. I determined that the limit is 1 > 0, which suggests that the series diverges because 1/n diverges. However, the doctor clarified why this test doesn't apply to the alternating sum.
Click to expand...
I didn't understand where the signs went in post #6. As Riemann's theorem tells us, we have to be very careful with the arrangement of the terms. You can only make comparisons term by term in alternating series.
 
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