Convergence of Series: Check a) & b)

[Mutaja]

Homework Statement

Check if the series below converge.

a) $$\sum_{n = 1}^\infty \frac{n}{2n^2 - 1}$$

b) $$\sum_{n = 2}^\infty (-1)^n \frac{2n}{n^2 - 1}$$

Homework Equations

The Attempt at a Solution

For a).

The series converge if the sum comes up to a finite value. If not, and it goes to positive or negative infinity, it diverges. That's all good.

I've also learned that if the common ratio is below 1 and greater than 0, the series will converge. The common ratio, as I understand it, is the expression that's containing n.

For instance, if the series is $2^n$, then the common ratio is 2. If you have a more advanced expression, and you obtain a constant, that constant is not part of the common ratio.

What I'm confused with here, I believe, is basic algebra. Is there a way I can re-write my expression ($\frac{n}{2n^2 - 1}$) to obtain the common ratio?

 

[jbunniii]

By "common ratio" I presume you mean $a_{n+1}/a_n$ where $a_n$ is the $n$'th term of the series $\sum_{n=1}^{\infty}a_n$.

For most series, $a_{n+1}/a_n$ is not constant for all $n$. In that case, you evaluate the limit:
$$\lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$$
If this limit exists and is less than 1, the series converges. If the limit exists and is greater than 1, it diverges. If the limit does not exist or equals 1, the test is inconclusive.

In your case, $a_n = n/(2n^2-1)$. What is $a_{n+1}$?

 

[jbunniii]

P.S. For part (a), it will probably be more useful to try the comparison test.

 

[SammyS]

Homework Statement

Check if the series below converge.

a) $$\sum_{n = 1}^\infty \frac{n}{2n^2 - 1}$$

b) $$\sum_{n = 2}^\infty (-1)^n \frac{2n}{n^2 - 1}$$

Homework Equations

The Attempt at a Solution

For a).

The series converge if the sum comes up to a finite value. If not, and it goes to positive or negative infinity, it diverges. That's all good.

I've also learned that if the common ratio is below 1 and greater than 0, the series will converge. The common ratio, as I understand it, is the expression that's containing n.

For instance, if the series is $2^n$, then the common ratio is 2. If you have a more advanced expression, and you obtain a constant, that constant is not part of the common ratio.

What I'm confused with here, I believe, is basic algebra. Is there a way I can re-write my expression ($\frac{n}{2n^2 - 1}$) to obtain the common ratio?

If $\displaystyle a_n=\frac{n}{2n^2 - 1}$, the what is $\ a_{n+1} \ \ ?$

Rather than this, divide the numerator & denominator by n and take the limit as n → ∞ .

 

[jbunniii]

Rather than this, divide the numerator & denominator by n and take the limit as n → ∞ .

I don't see how that will help in this case.

 

[SammyS]

I don't see how that will help in this case.

p-series

 

[jbunniii]

p-series

I agree with dividing numerator and denominator by $n$, but taking the limit as $n \rightarrow \infty$ will just give you zero, which isn't enough to conclude anything. Instead, I would apply the comparison test with an appropriate p-series.

 

[Mutaja]

P.S. For part (a), it will probably be more useful to try the comparison test.

Thanks a lot for your help, both of you. I really appreciate it.

For a) I've done the following:

$$\sum_{n = 1}^\infty \frac{n}{2n^2 - 1}$$ -> $$\frac{n}{2n^2}$$ since -1 is negligible (negligible?).

Dividing both the numerator and denominator by n, we get $$\frac{1}{2n}$$ which converges.

Therefore, $$\sum_{n = 1}^\infty \frac{n}{2n^2 - 1}$$ will converge.

This seems a bit too simple to me.

I've attempted b) as well, but let's focus on a) first. I want to get this right.

 

[HallsofIvy]

Thanks a lot for your help, both of you. I really appreciate it.

For a) I've done the following:

$$\sum_{n = 1}^\infty \frac{n}{2n^2 - 1}$$ -> $$\frac{n}{2n^2}$$ since -1 is negligible (negligible?).

Dividing both the numerator and denominator by n, we get $$\frac{1}{2n}$$ which converges.

You were fine up until you r last statement but, no, $\frac{1}{2}\sum\frac{1}{n}$ does NOT converge!

Therefore, $$\sum_{n = 1}^\infty \frac{n}{2n^2 - 1}$$ will converge.

This seems a bit too simple to me.

It is simple, but you still have to know what you are doing! SammyS suggested "p-series". What do you know about that?

I've attempted b) as well, but let's focus on a) first. I want to get this right.[/QUOTE]
What do you know about "alternating series"?

 

[Mutaja]

You were fine up until you r last statement but, no, $\frac{1}{2}\sum\frac{1}{n}$ does NOT converge!

Eek. I thought that if you add the series, and their sum is infinity, positive or negative, then it diverges. If you get a finite number, it converges? Since $\frac{1}{n}$ approaches 0 as n approaches infinity, then it will be a finite number? I'm obviously wrong here, but that's my thinking.

It is simple, but you still have to know what you are doing! SammyS suggested "p-series". What do you know about that?

I think that p-series is a series $\frac{1}{n^x}$ where for all x larger than 1, it converges. Equal or less than 1 it diverges. But I didn't think I had a series like that until now, when you took the constant outside of the series.

But regardless of that - if I had thought about it myself, I would be confused by the two methods. My ratio test - which is wrong, and showing that I have a p series with x=1.

What do you know about "alternating series"?

Series that contains $(-1)^n$ which will alternate the +/- sign between each part.

 

[HallsofIvy]

Eek. I thought that if you add the series, and their sum is infinity, positive or negative, then it diverges. If you get a finite number, it converges? Since $\frac{1}{n}$ approaches 0 as n approaches infinity, then it will be a finite number? I'm obviously wrong here, but that's my thinking.

Yes, your thinking is completely wrong! the terms approaching 0 is necessary for a series to converge but is NOT sufficient. That is, if the terms do NOT go to 0, you know immediately that the series diverges. If the term do go to 0, you still do not know if it diverges or converges.

I think that p-series is a series $\frac{1}{n^x}$ where for all x larger than 1, it converges. Equal or less than 1 it diverges. But I didn't think I had a series like that until now, when you took the constant outside of the series.

Surely you know that $c\sum a_n$ converges if and only if $\sum ca_n$ converges? You can always factor a constant out of a series.

But regardless of that - if I had thought about it myself, I would be confused by the two methods. My ratio test - which is wrong, and showing that I have a p series with x=1.

You said before that your ratio test was "inconclusive" didn't you? You said "I've also learned that if the common ratio is below 1 and greater than 0, the series will converge." But that does NOT mean that if the common ratio is not in that interval, the series will diverge- ratio= 1 is "inconclusive".

By the way, you also said, "If not, and it goes to positive or negative infinity, it diverges." Do you understand that a series may "diverge" but NOT go to positive or negative infinity? The series $\sum_{n=0}^\infty (-1)^n= 1- 1+ 1- 1+ \cdot\cdot\cdot$ does not converge (so "diverges") but does not go to either positive infinity or negative infinity.



Series that contains $(-1)^n$ which will alternate the +/- sign between each part.[/QUOTE]

 

[Mutaja]

Yes, your thinking is completely wrong! the terms approaching 0 is necessary for a series to converge but is NOT sufficient. That is, if the terms do NOT go to 0, you know immediately that the series diverges. If the term do go to 0, you still do not know if it diverges or converges.

How do I proceed then, if my comparison test is inconclusive? Let's leave my b) problem out of this for now. For my a) problem, jbunniii was kind enough to suggest to try the comparison test. My result from the comparison test is that it's inconclusive since $$\lim_{n => ∞}\frac{1}{2n}$$ approaches 0.

I'm supposed to check if the series converge, and my test is inconclusive. Is this my final answer, or can I use another method that's conclusive when the comparison test isn't?

Surely you know that $c\sum a_n$ converges if and only if $\sum ca_n$ converges? You can always factor a constant out of a series.

Yes, I do know that, but I wasn't thinking of it straight away. I need more training with these problems for that to come naturally for me, and I'll work on it.

You said before that your ratio test was "inconclusive" didn't you? You said "I've also learned that if the common ratio is below 1 and greater than 0, the series will converge." But that does NOT mean that if the common ratio is not in that interval, the series will diverge- ratio= 1 is "inconclusive".

I'm actually confused now as to how I did my ratio test. I've made an error when simplifying my expression. I struggle with simple algebra because we go months without using it. Sadly I haven't prioritized working on it myself - but I will do something about that now.

As for my aforementioned question, is the ratio test something I can do if my comparison test is inconclusive? If so, I will gladly post my work on the ratio test for problem a).

By the way, you also said, "If not, and it goes to positive or negative infinity, it diverges." Do you understand that a series may "diverge" but NOT go to positive or negative infinity? The series $\sum_{n=0}^\infty (-1)^n= 1- 1+ 1- 1+ \cdot\cdot\cdot$ does not converge (so "diverges") but does not go to either positive infinity or negative infinity.

I did not, because I have written that as a rule in my notes. I will change that, for sure.

Thanks a lot for helping me out, Sir. I will do my best to get this right, but as of now, I'm still not sure how to proceed, but I'm hoping that an answer to my question above will help me out.

I really appreciate that you're taking your time to help me.

 

[Ray Vickson]

I agree with dividing numerator and denominator by $n$, but taking the limit as $n \rightarrow \infty$ will just give you zero, which isn't enough to conclude anything. Instead, I would apply the comparison test with an appropriate p-series.

Taking $n \to \infty$ will show that for large $n$ the $n$th term behaves like $1/(2n)$ in the first case and like $2 (-1)^n /n$ in the second case.

 

[Mutaja]

Taking $n \to \infty$ will show that for large $n$ the $n$th term behaves like $1/(2n)$ in the first case and like $2 (-1)^n /n$ in the second case.

Will this be useful for the ratio test?

There are so many methods. How do I decide which one to use, what what to do if it's inconclusive?

 

[pasmith]

Homework Statement

Check if the series below converge.

a) $$\sum_{n = 1}^\infty \frac{n}{2n^2 - 1}$$

This seems to be a candidate for the integral test.

 

[Mutaja]

This seems to be a candidate for the integral test.

Alright, I'll try this method as well. I need all the repetition (?) I can get.

$$\sum_{n = 1}^\infty \frac{n}{2n^2 - 1}$$ -> $$\int_{1}^\infty \frac{n}{2n^2 - 1} dx$$

I need to make sure that the expression is positive: For x > 1, $\frac{n}{2n^2 - 1}$ > 0.

I also need to make sure that the series is decreasing. Since -1 is neglectable, $\frac{x}{2x^2}$ is clearly decreasing as x approaches infinity.

So to my integral.

$$\int_{1}^\infty \frac{n}{2n^2 - 1} dx$$

u = $2x^2 -1$ -> dx = $\frac{du}{4x}$

$$\int_{1}^\infty \frac{x}{u} * \frac{du}{4x}$$ -> $$\int_{1}^\infty \frac{1}{4u} du$$

Taking the constant outside of the integral: $\frac{1}{4}$ $$\int_{1}^\infty \frac{1}{u} du$$

Now I have $\frac{1}{4}$ [lnu] -> $\frac{1}{4}$ [ln 2x^2 - 1]

$\frac{1}{4}$ $ln(2t^2 - 1) - ln(2 - 1)$

Since ln(2 - 1) is = 0 what I'm left with is $\frac{1}{4}$ $ln(2t^2 - 1)$.

Since, again, -1 is neglectible, I have $\frac{1}{4}$ $$\lim_{t -> ∞} $$ $ln(2t^2 - 1)$

Which clearly is infinity. Thus, the original series diverges. How does this work look?

I tried to make it look good, but my latex codes aren't perfect. If someone could tell me why it's dropping down every time I use the $$ code, that would be great. Also, how do I get the [ ] brackets with the limits for integrals?

 

[jbunniii]

How do I proceed then, if my comparison test is inconclusive? Let's leave my b) problem out of this for now. For my a) problem, jbunniii was kind enough to suggest to try the comparison test. My result from the comparison test is that it's inconclusive since $$\lim_{n => ∞}\frac{1}{2n}$$ approaches 0.

I don't think you have used the comparison test correctly here. Suppose we wish to test the convergence of $\sum_n a_n$. For convenience we'll assume that $a_n \geq 0$ for all $n$ as is the case in this problem. Suppose I can find another series $\sum_{n}b_n$ (with $b_n \geq 0$), which is known to diverge, and suppose I can establish that $a_n \geq b_n$ for all $n$ (or, more generally, for all sufficiently large $n$). Then it follows that $\sum_{n} a_n$ must diverge.

Roughly speaking, $\sum_n b_n = \infty$ and $\sum_n a_n$ is "bigger" so it must also be $\infty$.

So apply this to your problem: note that $2n^2 - 1 < 2n^2$ for all $n$, so
$$\frac{1}{2n^2 - 1} > \frac{1}{2n^2}$$
for all $n$. Multiplying both sides by the positive number $n$, we get
$$\frac{n}{2n^2 - 1} > \frac{n}{2n^2} = \frac{1}{2n}$$
Thus *if* we know that $\sum_{n} 1/(2n)$ diverges, then $\sum_{n} n/(2n^2 - 1)$ must also diverge.

Reading your later posts, I can see that you didn't realize that
$$\sum_{n=1}^{\infty} \frac{1}{n}$$
diverges. The terms get smaller and smaller, but they don't get small fast enough for the sum to converge to a finite number. The proof of this is quite simple. Write out the first few terms of the sum:
$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \cdots$$
Let's group the terms together. The first group will have 1 term, the second group will have 2 terms, the third group will have 4 terms, and in general, the n'th group will have $2^{n-1}$ terms:
$$1 + \left(\frac{1}{2} + \frac{1}{3}\right) + \left(\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}\right) + \cdots$$
Now we'll apply the comparison test as follows:
$$1 + \left(\frac{1}{2} + \frac{1}{3}\right) + \left(\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}\right) + \cdots > 1 + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\right) + \cdots$$
On the right hand side, each group sums to $1/2$, and we are adding infinitely many of them. Thus the right hand side is $\infty$, and since the partial sums of the left side are even bigger, the sum is also $\infty$. So this proves that
$$\sum_{n=1}^{\infty} \frac{1}{n}$$
diverges. This is a good piece of information to commit to memory because it will be useful in comparison tests and other contexts. It's called the harmonic series, by the way:

http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

One interesting thing is that the harmonic series "barely" diverges. It crawls very slowly on its way to infinity. You have to add the first $10^{43}$ terms to even get to 100. Also, if instead of summing $\sum_{n} 1/n$, you sum $\sum_{n} 1/n^p$ where $p$ is any number greater than 1 (for example, 1.000000000000000001), the series converges!

 

[Mutaja]

One interesting thing is that the harmonic series "barely" diverges. It crawls very slowly on its way to infinity. You have to add the first $10^{43}$ terms to even get to 100. Also, if instead of summing $\sum_{n} 1/n$, you sum $\sum_{n} 1/n^p$ where $p$ is any number greater than 1 (for example, 1.000000000000000001), the series converges!

I will strive to reach your standards. Thanks doesn't justify how much I appreciate this response.

I will now try to solve my b) problem as you cleared up my confusion (at least for comparison tests) about the a) problem. I'd also like to note that, give that I've done my integral test correctly, it seems to me that it's a lot more work than the comparison test.

I will look into how the $(-1)^n$ affects the series in terms of convergence/divergence since it alternates the series. I will get back to this thread when I've done so.

Again, thank you so much.

 

[Ray Vickson]

I will strive to reach your standards. Thanks doesn't justify how much I appreciate this response.

I will now try to solve my b) problem as you cleared up my confusion (at least for comparison tests) about the a) problem. I'd also like to note that, give that I've done my integral test correctly, it seems to me that it's a lot more work than the comparison test.

I will look into how the $(-1)^n$ affects the series in terms of convergence/divergence since it alternates the series. I will get back to this thread when I've done so.

Again, thank you so much.

Google "alternating series".

BTW: a simpler proof of the divergence of $\sum 1/n$ is to look at the behavior of $S_N = \sum_{n=1}^N 1/n$ as $N \to \infty$. We have
$$1 > \int_1^2 \frac{dx}{x}, \: \frac{1}{2} > \int_2^3 \frac{dx}{x}, \ldots, \frac{1}{n} > \int_n^{n+1} \frac{dx}{x}$$
so
$$S_N = \sum_{n=1}^N \frac{1}{n} > \left( \int_1^2 + \int_2^3 + \cdots + \int_N^{N+1} \right) \frac{dx}{x} = \int_1^{N+1} \frac{dx}{x} = \ln(N+1).$$
Thus $S_N \to \infty$ as $N \to \infty$.

 

[jbunniii]

I will now try to solve my b) problem as you cleared up my confusion (at least for comparison tests) about the a) problem.

One thing to keep in mind about the comparison test is that it only applies directly to series with nonnegative terms. If you have negative terms, then you can use it to test for absolute convergence by considering the absolute values of the terms. But you can probably guess based on part (a) that the series
$$\sum_{n = 2}^\infty (-1)^n \frac{2n}{n^2 - 1}$$
is not absolutely convergent since the series of absolute values is
$$\sum_{n = 2}^\infty \frac{2n}{n^2 - 1}$$
which behaves similarly to the series in part (a).

So I would recommend a different test for (b). Do you know the alternating series test?

 

[jbunniii]

BTW: a simpler proof of the divergence of $\sum 1/n$

I'm not sure that's simpler. But I do like that approach too, because it shows why the sum diverges so slowly: the partial sums grow logarithmically. And also, comparing with integrals allows one to establish the convergence of $\sum 1/n^p$ for $p>1$, which is not so easy to do directly.

 

[Mutaja]

So I would recommend a different test for (b). Do you know the alternating series test?

I've attempted it, at least.

$$\sum_{n = 2}^\infty (-1)^n \frac{2n}{n^2 -1}$$

b_n = $\frac{2n}{n^2 -1}$

$$\lim_{n -> ∞} \frac{2n}{n^2 -1} = 0$$

b_n = $\frac{2n}{n^2 -1}$ > b_n+1 = $\frac{2(n+1)}{(n+1)^2 -1}$

Both conditions for the alternating series convergence test are met -> the series does converge.

The conditions are the limit of b_n as n approaches infinity is 0, and b_n is a decreasing sequence.

 

[Ray Vickson]

I'm not sure that's simpler. But I do like that approach too, because it shows why the sum diverges so slowly: the partial sums grow logarithmically. And also, comparing with integrals allows one to establish the convergence of $\sum 1/n^p$ for $p>1$, which is not so easy to do directly.

Right, and you can get an upper bound also:
$$\frac{1}{2} < \int_1^2 \frac{dx}{x}, \ldots, \frac{1}{n} < \int_{n-1}^n \frac{dx}{x}$$
so
$$S_N < 1 + \int_1^N \frac{dx}{x} = 1 + \ln(N)$$
Thus $\ln(N+1) < S_N < 1 + \ln(N)$. This works as well for p-series with p > 1.

 

[jbunniii]

$$\lim_{n -> ∞} \frac{2n}{n^2 -1} = 0$$

b_n = $\frac{2n}{n^2 -1}$ > b_n+1 = $\frac{2(n+1)}{(n+1)^2 -1}$

Both conditions for the alternating series convergence test are met -> the series does converge.

The conditions are the limit of b_n as n approaches infinity is 0, and b_n is a decreasing sequence.

Looks good to me. I would recommend that you show your work when claiming this inequality holds, since it is the key to the problem:
$$\frac{2n}{n^2 -1} > \frac{2(n+1)}{(n+1)^2 -1}$$
We can see that it's true since for large $n$, the left hand side looks like $2/n$ and the right hand side looks like $2/(n+1)$. Still, it's best to work it out explicitly, especially if it's being graded.

 

[Mutaja]

Looks good to me. I would recommend that you show your work when claiming this inequality holds, since it is the key to the problem:
$$\frac{2n}{n^2 -1} > \frac{2(n+1)}{(n+1)^2 -1}$$
We can see that it's true since for large $n$, the left hand side looks like $2/n$ and the right hand side looks like $2/(n+1)$. Still, it's best to work it out explicitly, especially if it's being graded.

I'm actually stuck at this. As I mentioned in one of my first posts in this thread, we do one thing at a time. Series and sequences was what we started with in January, then complex numbers, advanced matrices and MATLAB functions.

I can't remember how to solve such advanced inequalities. If there were one fraction, I would be fine - at least I would be able to make an attempt. Now I'm stuck thinking about what I do with the right hand side if I set $n^2 > 1$ for the left hand side.

I really want to do this, but my assignment is due in 4 hours and I still have work left. That's my problem, not yours, of course, but this thread won't be gone so I will look into it whenever I have time.

 

[SammyS]

I'm actually stuck at this. As I mentioned in one of my first posts in this thread, we do one thing at a time. Series and sequences was what we started with in January, then complex numbers, advanced matrices and MATLAB functions.

I can't remember how to solve such advanced inequalities. If there were one fraction, I would be fine - at least I would be able to make an attempt. Now I'm stuck thinking about what I do with the right hand side if I set $n^2 > 1$ for the left hand side.

I really want to do this, but my assignment is due in 4 hours and I still have work left. That's my problem, not yours, of course, but this thread won't be gone so I will look into it whenever I have time.

It's not all that advanced as inequalities go.

To show $\displaystyle\ \frac{n}{2n^2 - 1}\ge\frac{1}{2n}\,,\$ you can work "backwards to some inequality that you are confident is true. Then in your proof, reverse the order of the steps.

Starting with $\displaystyle\ \frac{n}{2n^2 - 1}\,,\$ divide the numerator & denominator by n .

Simplify this and factor out 2 from the denominator giving

$\displaystyle\ \frac{1}{2}\frac{1}{n - 1/(2n)}\ .\$​

How does this compare with $\displaystyle\ \frac{1}{2} \frac{1}{n}\ ?\$

They both have the same numerator. Compare their denominators:

Isn't $\ n-1/(2n) \$ less than $\ n \ ?\$ . Also they are both positive.

Therefore, $\displaystyle\ \frac{1}{n - 1/(2n)}>\frac{1}{n}\ .\$ Right?

 

[Ray Vickson]

I'm actually stuck at this. As I mentioned in one of my first posts in this thread, we do one thing at a time. Series and sequences was what we started with in January, then complex numbers, advanced matrices and MATLAB functions.

I can't remember how to solve such advanced inequalities. If there were one fraction, I would be fine - at least I would be able to make an attempt. Now I'm stuck thinking about what I do with the right hand side if I set $n^2 > 1$ for the left hand side.

I really want to do this, but my assignment is due in 4 hours and I still have work left. That's my problem, not yours, of course, but this thread won't be gone so I will look into it whenever I have time.

Look at the function $f(x) = x/(2x^2-1)$. Its derivative is $f'(x) = -(2x^2+1)/(2x^2-1)^2 < 0$ for $x > 1$. Therefore $f(n+1) < f(n), \; n=2, 3, 4, \ldots.$