Absolute Convergence Theorem and Test for Divergence Connection

[knowLittle]

Homework Statement

Determine whether the series is absolutely convergent, conditionally convergent, or divergent.

$\sum _{n=1}\left( -1\right) ^{n}\dfrac {n} {n^{2}+1}$ the sum goes to infinity.

Homework Equations

Theorem for absolute convergence.
Test for divergence

The Attempt at a Solution

By the theorem of Absolute convergence: If the absolute value of the sum converges, then the sum converges.

$\sum _{n=1}\left| \left( -1\right) ^{n}\dfrac {n} {n^{2}+1}\right| =\sum \dfrac {\dfrac {1} {n}} {1+\dfrac {1} {n^{2}}}=0$

I found that it converges to zero. Now, If this was the test for divergence, the test would be inconclusive. Since it is the absolute convergence theorem, the sum without the absolute value also converges.

Is this reasoning correct?

Thank you.

 

[LCKurtz]

Homework Statement

Determine whether the series is absolutely convergent, conditionally convergent, or divergent.

$\sum _{n=1}\left( -1\right) ^{n}\dfrac {n} {n^{2}+1}$ the sum goes to infinity.

Homework Equations

Theorem for absolute convergence.
Test for divergence

The Attempt at a Solution

By the theorem of Absolute convergence: If the absolute value of the sum converges, then the sum converges.

$\sum _{n=1}\left| \left( -1\right) ^{n}\dfrac {n} {n^{2}+1}\right| =\sum \dfrac {\dfrac {1} {n}} {1+\dfrac {1} {n^{2}}}=0$

I found that it converges to zero.

That's absurd on the face of it. All the terms in your absolute value series are positive. How can they possibly add up to 0?

Now, If this was the test for divergence, the test would be inconclusive. Since it is the absolute convergence theorem, the sum without the absolute value also converges.

Is this reasoning correct?

Thank you.

You haven't settled the absolute value series yet. If your absolute value series converges, then yes, the original series would be convergent. But you have a ways to go here. You haven't established anything yet. You might consider a comparison test on the absolute value series to get started.

 

[knowLittle]

I don't know how to settle the absolute value series.
However, I did some limit comparison of the original series with the alternating harmonic and found that both converge.

Then, I used the limit comparison test of the absolute value series with a divergent harmonic and found that both diverge.

Can I say that the infinite series converges conditionally?

 

[LCKurtz]

I don't know how to settle the absolute value series.
However, I did some limit comparison of the original series with the alternating harmonic and found that both converge.

I have never heard of any kind of limit comparison test for alternating series, so I doubt whatever you did was correct, whether or not the result was.

Then, I used the limit comparison test of the absolute value series with a divergent harmonic and found that both diverge.

That may well be correct, but from what I haves seen of your stuff, I would want to see your work before I said so.

Can I say that the infinite series converges conditionally?

That would be correct if your conclusions are.

 

[knowLittle]

I have never heard of any kind of limit comparison test for alternating series, so I doubt whatever you did was correct, whether or not the result was.

You are right. I can't use the limit comparison with alternating series, since the limit comparison can only be done when the terms compared are greater than 0.

That may well be correct, but from what I haves seen of your stuff, I would want to see your work before I said so.

Here it is:
$\lim _{n\rightarrow \infty }\left[ \dfrac {\dfrac {n} {n^{2}+1}} {\dfrac {1} {n}}=\dfrac {n^{2}} {n^{2}+1}=\dfrac {1} {1+\dfrac {1} {n^{2}}}\right] =1$
Where the limit comparison test tells us that if the limit >0 and is finite. Then, either both series converge or both diverge. I used the divergent harmonic series; therefore, the other series also diverge.

 

[LCKurtz]

You are right. I can't use the limit comparison with alternating series, since the limit comparison can only be done when the terms compared are greater than 0.




Here it is:
$\lim _{n\rightarrow \infty }\left[ \dfrac {\dfrac {n} {n^{2}+1}} {\dfrac {1} {n}}=\dfrac {n^{2}} {n^{2}+1}=\dfrac {1} {1+\dfrac {1} {n^{2}}}\right] =1$
Where the limit comparison test tells us that if the limit >0 and is finite. Then, either both series converge or both diverge. I used the divergent harmonic series; therefore, the other series also diverge.

Giood, that's correct. So you know the original series is not absolutely convergent. So you have yet to establish whether it is conditionally convergent.

 

[knowLittle]

Ok. So, I can't use the alternating series test. I will use the ratio test. I got an inconclusive ratio test.

I don't know how to proceed.

 

[LCKurtz]

Ok. So, I can't use the alternating series test. I will use the ratio test. I got an inconclusive ratio test.

I don't know how to proceed.

The ratio test tests for absolute convergence and you already know it isn't absolutely convergent.

Why do you think you can't use the alternating series test? It is an alternating series after all.

 

[knowLittle]

You are right.
I can use the alternating series test. What I can't use is the limit test with an alternating series, since for that limit test both values have to be >0.

So, for the alternating series test, I found as the lim approaches infinity of term an
$\lim _{n\rightarrow \infty }[\dfrac {n} {n^{2}+1}=\left( \dfrac {\dfrac {1} {n}} {1+\dfrac {1} {n^{2}}}\right) =0$
I don't know how to show that A_(n+1) < A_n

Can I find the derivative of An and as in the integral test find, if the function is decreasing?

 

[knowLittle]

Finally,
$f^{1}\left( x\right) =\dfrac {-\left( x^{2}-1\right) } {\left( x^{2}+1\right) ^{2}}$
for any value >=1, the derivative is negative and therefore decreasing.

Is this argument correct?
Can I conclude that by the alternating series test my series converges. Then, the series converges conditionally.

 

[LCKurtz]

Finally,
$f^{1}\left( x\right) =\dfrac {-\left( x^{2}-1\right) } {\left( x^{2}+1\right) ^{2}}$
for any value >=1, the derivative is negative and therefore decreasing.

Is this argument correct?
Can I conclude that by the alternating series test my series converges. Then, the series converges conditionally.

Yes. That is a good argument. You could also show$$
\frac{n+1}{(n+1)^2+1}\le \frac n {n^2+1}$$directly by simplifying it with algebra.

 

[knowLittle]

Yes. That is a good argument. You could also show$$
\frac{n+1}{(n+1)^2+1}\le \frac n {n^2+1}$$directly by simplifying it with algebra.

I'm having trouble in showing it through Algebra.

I know that (n^2)+1 < (n+1)^2 +1 ; then, 1/( (n+1)^2 +1 ) < 1/( (n^2) +1)
Then, I could multiply both sides by n/ ( (n+1)^2 +1 ) < n/ ( (n^2) +1).

How do I show the last step in the numerator?

 

[Dick]

I'm having trouble in showing it through Algebra.

I know that (n^2)+1 < (n+1)^2 +1 ; then, 1/( (n+1)^2 +1 ) < 1/( (n^2) +1)
Then, I could multiply both sides by n/ ( (n+1)^2 +1 ) < n/ ( (n^2) +1).

How do I show the last step in the numerator?

You can't show it through algebra because it's not true. It's false for n=0. It's only true for n>1. I like the derivative argument better.

 

[LCKurtz]

Yes. That is a good argument. You could also show$$
\frac{n+1}{(n+1)^2+1}\le \frac n {n^2+1}$$directly by simplifying it with algebra.

You can't show it through algebra because it's not true. It's false for n=0. It's only true for n>1. I like the derivative argument better.

Of course you can show it with algebra and it's only necessary for it to be true for large $n$. But in fact it is true for $n\ge 1$ and straightforward. Cross multiply (I switched the sides, I don't know why but I don't want to retype it)$$
n((n+1)^2 + 1) \ge (n+1)(n^2+1)$$ $$
n(n^2+2n+2)\ge n^3+n^2+n+1$$ $$
n^3+2n^2+2n\ge n^3+n^2+n+1$$ $$
n^2+n\ge 1$$All the steps are reversible. Beauty is in the eye of the beholder; I would have done it this way.

 

[knowLittle]

It looks good. Thank you.

 

[LCKurtz]

It looks good. Thank you.

You're welcome. I hope you feel like you learned something from our exchanges today.

 

[knowLittle]

I did. :)