Limit of a_n as n Approaches Infinity & Zero

[agnimusayoti]

Homework Statement

Use the preliminary test to decide the following series are divergent.
$$\frac{1}{2}-\frac{4}{5}+\frac{9}{10}-\frac{16}{17}+\frac{25}{26}+...$$

Relevant Equations

Preliminary test: if $\lim_{n\to\infty} a_n = 0$ then the series $\sum_1^\infty{a_n}$ is a divergent series

1. From the series, I got that $a_n=\frac{(-1)^{n+1} n^2}{n^2+1}$. I think my answer is correct. But if I try to find the limit as n approaching infinity, I got indeterminate form (isn't it?) : $\frac{-1^\infty}{1+0}$
2. What is the limit as n approaching zero?

Thanks, all!

 

[haruspex]

Homework Statement:: Use the preliminary test to decide the following series are divergent.
$$\frac{1}{2}-\frac{4}{5}+\frac{9}{10}-\frac{16}{17}+\frac{25}{26}+...$$
Relevant Equations:: Preliminary test: if $\lim_{n\to\infty} a_n = 0$ then the series $\sum_1^\infty{a_n}$ is a divergent series

1. From the series, I got that $a_n=\frac{(-1)^{n+1} n^2}{n^2+1}$. I think my answer is correct. But if I try to find the limit as n approaching infinity, I got indeterminate form (isn't it?) : $\frac{-1^\infty}{1+0}$
2. What is the limit as n approaching zero?

Thanks, all!

I think you mean: Preliminary test: if $\lim_{n\to\infty} a_n \neq 0$ then the series $\sum_1^\infty{a_n}$ is a divergent series. But even that is not quite right. See e.g. http://sites.science.oregonstate.ed...StudyGuides/SandS/SeriesTests/divergence.html.

 

[Mark44]

Homework Statement:: Use the preliminary test to decide the following series are divergent.
$$\frac{1}{2}-\frac{4}{5}+\frac{9}{10}-\frac{16}{17}+\frac{25}{26}+...$$
Relevant Equations:: Preliminary test: if $\lim_{n\to\infty} a_n = 0$ then the series $\sum_1^\infty{a_n}$ is a divergent series

1. From the series, I got that $a_n=\frac{(-1)^{n+1} n^2}{n^2+1}$. I think my answer is correct.

Under Relevant Equations, the test you gave is for series with positive terms. The series you're working with is an alternating series, which generally look like $\sum (-1)^n a_n$ . Here $a_n = \frac{n^2}{n^2 + 1}$. What is $\lim_{n \to \infty}\frac {n^2}{n^2 + 1}$?

But if I try to find the limit as n approaching infinity, I got indeterminate form (isn't it?) : $\frac{-1^\infty}{1+0}$

2. What is the limit as n approaching zero?

Why? The limit as n approaches zero (not n approaching zero) is completely unrelated to this problem.

 

[haruspex]

Under Relevant Equations, the test you gave is for series with positive terms.

No, it’s a sufficiency test for non-convergence. The error was in writing $=$ instead of $\neq$.

 

[agnimusayoti]

Uh yeah. I made a mistake in preliminary test definition. What I mean is, $\lim_{n\to\infty} a_n\neq0$.

 

[agnimusayoti]

Under Relevant Equations, the test you gave is for series with positive terms. The series you're working with is an alternating series, which generally look like $\sum (-1)^n a_n$ . Here $a_n = \frac{n^2}{n^2 + 1}$. What is $\lim_{n \to \infty}\frac {n^2}{n^2 + 1}$?
Why? The limit as n approaches zero (not n approaching zero) is completely unrelated to this problem.

1. I made a mistake again in typing question number 2 (TT). I mean as n approaching infinity.
2. I still don't get why $(-1)^n$ is not included in $a_n$ formula.
3. That limit, $\lim_{n\to\infty} \frac{n^2}{n^2+1}=\frac{1}{1+0}=1$

 

[agnimusayoti]

What I confuse is the result of $(-1)^\infty$. I think $(-1)^\infty$ is indeterminate form. Or may be I don't have sufficient concept about indeterminate form?

 

[haruspex]

What I confuse is the result of $(-1)^\infty$. I think $(-1)^\infty$ is indeterminate form. Or may be I don't have sufficient concept about indeterminate form?

Did you follow the link I posted in #2? Do you see the other difference between what you quoted as the test and what that link quotes? I don't mean the equals/not equals - the other difference.

 

[agnimusayoti]

Yes, I follow your link. Do you mean the limit doesn't exist so the series is divergent?

 

[agnimusayoti]

I think I need to re-evaluate... May be I can use l'Hopital's theorem...

 

[haruspex]

Yes, I follow your link. Do you mean the limit doesn't exist so the series is divergent?

Yes. Haven't you shown the sequence does not converge to a limit?

 

[agnimusayoti]

Yes. Haven't you shown the sequence does not converge to a limit?

mm, the sequence tends to 1 or (-1). I'm not sure about it... But as n getting bigger, odd term is approaching 1; even term is approaching -1. Is it right? So it is not necessary to compute the limit (because the sequence itself does not converge to a limit), isn't it?
But, how do you know the limit doesn't exist if we did not try to compute the limit?

If I try to use l'Hopital's theorem, I use $f(n)=(-1)^n n^2 \rightarrow f'(n)=2n(-1)^n+(-1)^n \ln (-1) n^2$. But, $\ln(-1)$is error and $g(n)=(n+1)^2 \rightarrow g'(n)=2(n+1)(1)=2(n+1)$. So, because ln(-1) doesn't exist the theorem's can not be used and therefore the limit does not exist. Is it right?

 

[haruspex]

the sequence itself does not converge to a limit

Quite so. To put it another way, the limit of the sequence does not exist. So according to the "preliminary" test, the series does not converge either.

 

[agnimusayoti]

Then, is every "alternating sign" series always divergent series? Thanks again.

 

[haruspex]

Then, is every "alternating sign" series always divergent series? Thanks again.

No. If the absolute value of the terms tends to zero then the sequence tends to zero whether the signs alternate or not. In such a case, the preliminary test doesn't tell you whether the series converges.

 

[agnimusayoti]

No. If the absolute value of the terms tends to zero then the sequence tends to zero whether the signs alternate or not. In such a case, the preliminary test doesn't tell you whether the series converges.

Oh, therefore I need further test if the sequence tends to a certain, non zero limit.

 

[haruspex]

Oh, therefore I need further test if the sequence tends to a certain, non zero limit.

Not certain what you mean.

If the sequence does not converge to a limit then the series does not converge. That is the situation in this thread.

If the sequence does converge to a limit but that limit is not zero then the series does not converge.

It doesn’t matter whether you consider the terms including their signs or take their absolute values. In this thread, the absolute values do converge to 1, so the first part of the test allows that the series might converge, but this doesn't matter because the divergence will be detected by the second part of the test.

 

[agnimusayoti]

Under Relevant Equations, the test you gave is for series with positive terms. The series you're working with is an alternating series, which generally look like $\sum (-1)^n a_n$ . Here $a_n = \frac{n^2}{n^2 + 1}$. What is $\lim_{n \to \infty}\frac {n^2}{n^2 + 1}$?
Why? The limit as n approaches zero (not n approaching zero) is completely unrelated to this problem.

Could you explain why the test is for series with positive terms? Because the book does not show it . Thankss

 

[Mark44]

Could you explain why the test is for series with positive terms? Because the book does not show it . Thankss

I misspoke in what I said about the Nth Term Test for Divergence, which is what you had.
However, what I said about alternating series is correct; namely, that they have the form $\sum (-1)^n a_n$, with $a_n > 0$.

 

[Mark44]

Then, is every "alternating sign" series always divergent series? Thanks again.

No. Some alternating series converge, and some diverge.
For example, the series $\sum_{n = 1}^\infty (-1)^{n+1}\frac 1 n$ is a convergent series, and is known as the alternating harmonic series. On the other hand, the series $\sum_{n = 1}^\infty (-1)^n n$ is a divergent alternating series.

 

[agnimusayoti]

I misspoke in what I said about the Nth Term Test for Divergence, which is what you had.
However, what I said about alternating series is correct; namely, that they have the form $\sum (-1)^n a_n$, with $a_n > 0$.

But, if I try to find the limit of positive terms, I find that if the limit is not equal to 0, then the series is divergent. ANd it is fit with the answer book. Because of this experience and what you say, I asked my self, whether I can use preliminary test just for the positive terms. If it so, I think it is easier than try to proof that the limit does not exist...

 

[haruspex]

But, if I try to find the limit of positive terms, I find that if the limit is not equal to 0, then the series is divergent. ANd it is fit with the answer book. Because of this experience and what you say, I asked my self, whether I can use preliminary test just for the positive terms. If it so, I think it is easier than try to proof that the limit does not exist...

Please read/reread my post #17.
For the purposes of the "preliminary test", it does not matter whether the series is of all positive terms, of alternating sign, or whatever mix of positive and negative terms; neither does it matter whether you apply the test to the terms with whatever sign they have or to the absolute value of the terms.

If the series has a mix of signs, some series that fail the test will fail at the first stage (the sequence does not converge) when you include the signs, but fail instead at the second stage (the sequence limit exists but is not zero) if you use the absolute values of the terms. But the same set of series will fail the test at some point whether you use the signed or unsigned values.

The signs do matter if the series passes the preliminary test, i.e., if the preliminary test does not rule out convergence of the series, so you need to move on to more subtle tests. In this case, it often happens that $\Sigma a_i$ converges but $\Sigma |a_i|$ does not.

 

[agnimusayoti]

neither does it matter whether you apply the test to the terms with whatever sign they have or to the absolute value of the terms.

So can I apply preliminary test to the absolute value of the terms? If that so, could you explain why it is possible?

If the series has a mix of signs, some series that fail the test will fail at the first stage (the sequence does not converge) when you include the signs, but fail instead at the second stage (the sequence limit exists but is not zero) if you use the absolute values of the terms. But the same set of series will fail the test at some point whether you use the signed or unsigned values.

I do not understand this part. Could you please explain again, especially on "first stage" and "second stage"?

 

[Mark44]

Many calculus textbooks introduce the topic of infinite series in this order:

• Sequences and how to determine whether they converge or diverge
• Series with only positive terms, such as $\sum \frac 1 {n^2}$ (a convergent series), and $\sum \frac 1 n$ (a divergent series)
• Alternating series of the form $\sum (-1)^{n+1}a_n$, with $a_n > 0$.
• Power series such as $\sum a_nx^n$

The series you showed in post #1 is an alternating series, to which you can apply the Alternating Series Test (see https://en.wikipedia.org/wiki/Alternating_series_test). This test says that, for a series of the form $\sum (-1)^{n+1}a_n$, with $a_n > 0$, the series converges if 1) the sequence $\{a_n\}$ decreases monotonically; i.e., $a_{n+1} < a_n$ for all n larger than some number N, and 2) $\lim_{n \to \infty}a_n = 0$.

If you apply the alternating series test to the series in this thread, you should be able to determine whether your series converges or diverges.

 

[haruspex]

So can I apply preliminary test to the absolute value of the terms? If that so, could you explain why it is possible?I do not understand this part. Could you please explain again, especially on "first stage" and "second stage"?

The test as prescribed at the link I gave is in two parts.
First part: if the sequence does not converge to a limit then the series does not converge.
Second part: if the sequence does converge to a limit but that limit is not zero then the series does not converge.

If you apply this test to the actual sequence, including the alternating sign, it falls at the first hurdle: the sequence does not converge.
If you apply this test to the unsigned sequence it passes the first test because |a_i| does converge, but then it falls at the second hurdle because it converges to 1, not zero.

What is less obvious is that this is a general behaviour of the test: if applying the test to a sequence a_i with varying signs shows the series does not converge then applying it to |a_i| will also show that the series does not converge, and vice versa. It does not matter which you choose.

This is because if |a_i| converges, to c say, but a_i does not converge then c is not zero.

The series you showed in post #1 is an alternating series, to which you can apply the Alternating Series Test (see https://en.wikipedia.org/wiki/Alternating_series_test). This test says that, for a series of the form $\sum (-1)^{n+1}a_n$, with $a_n > 0$, the series converges if 1) the sequence $\{a_n\}$ decreases monotonically; i.e., $a_{n+1} < a_n$ for all n larger than some number N, and 2) $\lim_{n \to \infty}a_n = 0$.

Just to be clear for the OP, the monotonicity part is a sufficient condition for convergence of the series, but not a necessary one. E.g. $\Sigma (-1)^n\frac 1n|\sin(n)|$ converges even though $\frac 1n|\sin(n)|$ never becomes monotone.

Also, there are not just positive sequences and alternating sequences. The sign can change in other patterns.