Comparison vs. limit comparison vs. sum splitting

[ktheo]

I'm not really sure when each of these should be done. In fact, I don't really understand the reason that we use the limit comparison test.

Σ1/(n^2+1)

So here I can simply say that P=2>1, so the original converges.

Σ1/N^3+N^2

Here, I would say that P=3>1, implying the original converges. But my solutions tell me that here I should use a limit comparison with 1/N^3, where the limit ---->N is = 1. So why did I do this? What did I prove by finding the limit that I didn't prove by just comparing directly? What is different in this that I had to use a limit comparison instead of direct? Just the added N^2? I don't understand why. At first I thought it had to do with fulfilling the inequality where An<Bn of the two series if it converges, but in this case, the 1/n^3+n^2<1^n3, so I'm not sure.

And I also see that when I have some questions that have similar sums to the one directly above me, they just split the sum and evaluate the p series of both sums to find the divergence; Why does this seem to be an alternative?

Thank you, guys.

 

[jbunniii]

I'm not really sure when each of these should be done. In fact, I don't really understand the reason that we use the limit comparison test.

Σ1/(n^2+1)

So here I can simply say that P=2>1, so the original converges.

To be more precise, $0 < n^2 < n^2 + 1$, so
$$0 < \frac{1}{n^2 + 1} < \frac{1}{n^2}$$
so the series converges by comparison with $\sum 1/n^2$

Σ1/N^3+N^2

Here, I would say that P=3>1, implying the original converges.

You didn't use parentheses, but I assume you mean
$$\sum_{n=1}^{\infty} \frac{1}{n^3 + n^2}$$
If so, then we can compare with either $1/n^3$ or $1/n^2$ as follows:
$$n^3 + n^2 > n^3 > 0$$
so
$$0 < \frac{1}{n^3 + n^2} < \frac{1}{n^3}$$
Therefore the series converges by comparison with $\sum 1/n^3$. Similarly,
$$n^3 + n^2 > n^2 > 0$$
so
$$0 < \frac{1}{n^3 + n^2} < \frac{1}{n^2}$$
so the series converges by comparison with $\sum 1/n^2$.

 

[ktheo]

To be more precise, $0 < n^2 < n^2 + 1$, so
$$0 < \frac{1}{n^2 + 1} < \frac{1}{n^2}$$
so the series converges by comparison with $\sum 1/n^2$You didn't use parentheses, but I assume you mean
$$\sum_{n=1}^{\infty} \frac{1}{n^3 + n^2}$$
If so, then we can compare with either $1/n^3$ or $1/n^2$ as follows:
$$n^3 + n^2 > n^3 > 0$$
so
$$0 < \frac{1}{n^3 + n^2} < \frac{1}{n^3}$$
Therefore the series converges by comparison with $\sum 1/n^3$. Similarly,
$$n^3 + n^2 > n^2 > 0$$
so
$$0 < \frac{1}{n^3 + n^2} < \frac{1}{n^2}$$
so the series converges by comparison with $\sum 1/n^2$.

The way you just did it is exactly how I initially solved both problems... but on the second question, where I split the sums like you did, for the answer key, they used a limit comparison to prove the answer by stating something essentially equivalent to lim AN/BN = 1 using 1/(n^3) for Bn proving convergence by the limit comparison. I'm just not sure why they go to the effort of doing this when from what I can tell I can just split the sums and compare; is it just preference? Could you quickly sum up to me why I would consider using a LCT?

I had assumed that LCT was used only when you couldn't show the inequality like the one you wrote out for the first question I.E. An < Bn... but that seems to be an incorrect assessment.

 

[jbunniii]

The way you just did it is exactly how I initially solved both problems... but on the second question, where I split the sums like you did, for the answer key, they used a limit comparison to prove the answer by stating something essentially equivalent to lim AN/BN = 1 using 1/(n^3) for Bn proving convergence by the limit comparison. I'm just not sure why they go to the effort of doing this when from what I can tell I can just split the sums and compare; is it just preference? Could you quickly sum up to me why I would consider using a LCT?

I had assumed that LCT was used only when you couldn't show the inequality like the one you wrote out for the first question I.E. An < Bn... but that seems to be an incorrect assessment.

When trying to prove that a series converges, there are often several methods which will work. You can use whichever method you like in that case. For simple series like this, I like to use a direct comparison because I think it is clearest, but that's just my preference.

 

[arildno]

The point with exercises is not always to find A way that works, but rather, to polish your skills of using a specified way to do it, so that gradually, you become as used to picking THAT technique up from your toolbox as the examples who at present can be solved more easily with other techniques.

 

[ktheo]

Okay I understand. So given your preferences and looking at another question here... how would you choose to solve

Ʃ(sin)(1+/(n^2+n)

I would show sin alternates -1<sin<1 and compare on bn of Ʃ1/n^2 implying convergence of An, where An < Bn.

In his solutions, he does an extremely algebraically complicated limit comparison test... but if I can just solve via DC for full marks... I would assume he's just showing us this to offer the most in-depth answer?

 

[ktheo]

If I can take this one step further, would I be correct in saying that the main reason one would use the limit comparison test when it is difficult to establish the inequality using direct comparison?

 

[arildno]

You shouln't be given ANY marks at all, if the REQUIREMENT is that you should do it by limit comparison.

If, however, no such requirement is directly stated, you should gain full marks.

Thus, always read very carefully the requirements set up in the test.

 

[jbunniii]

Okay I understand. So given your preferences and looking at another question here... how would you choose to solve

Ʃ(sin)(1+/(n^2+n)

Do you mean
$$\sum_{n=1}^{\infty} \sin\left(\frac{1}{n^2 + n}\right)$$
or something else? If the above is correct, I would use the comparison
$$|\sin(x)| \leq |x|$$
so in particular
$$\left|\sin\left(\frac{1}{n^2 + n}\right)\right| \leq \frac{1}{n^2 + n} \leq \frac{1}{n^2}$$

 

[ktheo]

You shouln't be given ANY marks at all, if the REQUIREMENT is that you should do it by limit comparison.

If, however, no such requirement is directly stated, you should gain full marks.

Thus, always read very carefully the requirements set up in the test.

Okay thanks. This is the general question form I've seen in all past finals:

"For each of the following series, decide whether they converge or not. State
the tests you are using, and state and verify the hypotheses of the tests you are using."

I think this leaves me free to use whatever tool I like. Good to know. You've both been helpful thanks!

 

[ktheo]

Do you mean
$$\sum_{n=1}^{\infty} \sin\left(\frac{1}{n^2 + n}\right)$$
or something else? If the above is correct, I would use the comparison
$$|\sin(x)| \leq |x|$$
so in particular
$$\left|\sin\left(\frac{1}{n^2 + n}\right)\right| \leq \frac{1}{n^2 + n} \leq \frac{1}{n^2}$$

Yep, exactly this.

 

[jbunniii]

If I can take this one step further, would I be correct in saying that the main reason one would use the limit comparison test when it is difficult to establish the inequality using direct comparison?

I think that is a reasonable statement.

 

[arildno]

If I can take this one step further, would I be correct in saying that the main reason one would use the limit comparison test when it is difficult to establish the inequality using direct comparison?

Yup.

I don't remember all the details about these different tests, like the root test, limit comparison and testing against integrals, but the justification for learning them is to familiarize the student with lots of different weapons, and some of those weapons will seem unwieldy (but usable) in simple cases, while they are the ones you'll have to use when the problems get tougher.
And, then it will be very useful already to have in your arsenal the weapons you need.
Agreed?

 

[arildno]

Okay thanks. This is the general question form I've seen in all past finals:

"For each of the following series, decide whether they converge or not. State
the tests you are using, and state and verify the hypotheses of the tests you are using."

I think this leaves me free to use whatever tool I like. Good to know. You've both been helpful thanks!

Yes. That makes you free to use whatever tool you like.

 

[arildno]

NB!
Also remember that when you do not manage to decide convergence/divergence, your score can become much larger than zero if the format of answering allows you to show the reviewers precisely WHICH techniques you have tried out.

 

[ktheo]

NB!
Also remember that when you do not manage to decide convergence/divergence, your score can become much larger than zero if the format of answering allows you to show the reviewers precisely WHICH techniques you have tried out.

Appreciate the input for sure. The final exam is quickly approaching and I think I'm ready; I was just a little phased by the fact that essentially as I have been reviewing, I have been showing convergence or divergence ALWAYS using either the integral test, ratio test (which I clue into use whenever I have N powers that I can't convert to a geometric ratio), and the comparison test.

My confidence was just a little shaken on the midterm when we were given a question where it would be easiest to split the sums and do a p-series, but I did a limit comparison; I lost 4 of 5 marks because I said An≤Bn even though it wasn't true (which was why i did the LC in the first place; the inequality was difficult to obtain)... just thought it was weird I lost so much for that simple mistake, but I guess by putting that I ruined my answer since it seems as though convergence testing is much more intuitive and understanding then it is sheer mechanics.