Is my answer correct? (comparison test of series)

[Hn_zo]

Hello everyone, my Calculus III teacher sent us this exercise to show him sunday, and we need to show if the series is convergent or divergent using the comparison test. This is what I thought, and I'd like to know if it is correct or what I got wrong (he also said we need to do a divergent test, which I didn't include because I didn't know how to calculate the limit of the series, any help with this too would be great)

 

[mario99]

I think that your reasoning is correct because

[imath]\displaystyle \sum_{n=1}^{\infty}\frac{2^n \sin^2(5n)}{4^n + \cos^2(n)} \leq \sum_{n=1}^{\infty}\frac{2^n}{4^n}[/imath]

And

[imath]\displaystyle \sum_{n=1}^{\infty}\frac{2^n}{4^n}[/imath] is a convergent geometric series.

If you want to calculate the limit of the series do this:

[imath]\displaystyle \lim_{n\rightarrow \infty}\frac{2^n \sin^2(5n)}{4^n + \cos^2(n)} = \lim_{n\rightarrow \infty} e^{n\ln(2/4)} = \lim_{n\rightarrow \infty} e^{n\ln(1/2)} = \lim_{n\rightarrow \infty} e^{n[ \ \ln(1) - \ln(2) \ ]} = \lim_{n\rightarrow \infty} e^{n[ - \ln(2) \ ]} = e^{-\infty} = 0[/imath]

We ignored [imath]\displaystyle \sin^2(5n)[/imath] and [imath]\cos^2(n)[/imath] because they are just a number between [imath]0[/imath] and [imath]1[/imath].

 

[Dr.Peterson]

If you want to calculate the limit of the series do this:

No, that just says that the limit of the (sequence of) terms is zero, right? And that, in itself, doesn't even prove the series converges, only that it is possible.

 

[mario99]

No, that just says that the limit of the (sequence of) terms is zero, right? And that, in itself, doesn't even prove the series converges, only that it is possible.

Saying literally the limit of the series was just a slip of the tongue. I meant to say the limit of the sequence. And I think that the OP meant the same thing. I should have to reverse the order of solution. Check the limit first. When it is zero, then I have to prove the series converge. I did the reverse because it was too late when I realized that the OP needed help in the limit.

If the OP meant by the limit that he did not know how to calculate the exact result of the summation, I think that there is no easy way to do this in this specific series. If it were an alternating or telescoping series, I think that it would be possible.

So, Dr.Peterson, enlighten us, is there an easy way to find the exact limit of this series?

The only thing that I can say is that the sum of the original series must be [imath]> 0[/imath] and [imath]\displaystyle \leq \sum_{n=1}^{\infty}\left(\frac{1}{2}\right)^n[/imath].

 

[Dr.Peterson]

Perhaps I'm unusual, but I avoid the phrase "limit of a series"; what you are talking about are the sum of the series and the limit of the sequence. That was my main point: mixing the terms invites confusion.

To be honest, I hadn't noticed the OP's use of the phrase:

he also said we need to do a divergent test, which I didn't include because I didn't know how to calculate the limit of the series, any help with this too would be great

I suspect they are confused by the phrase, since the limit used in the divergence test is easy in this case.

But also, I see no need to separately do a divergence test, because comparison to a geometric series already incorporates that; if you can show by any means that a series converges, then its terms must converge to 0. Perhaps that is just this teacher's practice, in order to keep the divergence test on students' minds.

And, no, I don't see any way to determine (easily) the sum of the series, which is not required for the problem.

 

[mario99]

I suspect they are confused by the phrase, since the limit used in the divergence test is easy in this case.

And, no, I don't see any way to determine (easily) the sum of the series, which is not required for the problem.

That was exactly my thinking!

 

[Hn_zo]

I think that your reasoning is correct because

[imath]\displaystyle \sum_{n=1}^{\infty}\frac{2^n \sin^2(5n)}{4^n + \cos^2(n)} \leq \sum_{n=1}^{\infty}\frac{2^n}{4^n}[/imath]

And

[imath]\displaystyle \sum_{n=1}^{\infty}\frac{2^n}{4^n}[/imath] is a convergent geometric series.

If you want to calculate the limit of the series do this:

[imath]\displaystyle \lim_{n\rightarrow \infty}\frac{2^n \sin^2(5n)}{4^n + \cos^2(n)} = \lim_{n\rightarrow \infty} e^{n\ln(2/4)} = \lim_{n\rightarrow \infty} e^{n\ln(1/2)} = \lim_{n\rightarrow \infty} e^{n[ \ \ln(1) - \ln(2) \ ]} = \lim_{n\rightarrow \infty} e^{n[ - \ln(2) \ ]} = e^{-\infty} = 0[/imath]

We ignored [imath]\displaystyle \sin^2(5n)[/imath] and [imath]\cos^2(n)[/imath] because they are just a number between [imath]0[/imath] and [imath]1[/imath].

Hello, thank you for helping me. I was able to do the limit of the series using the confrontation test (the result was also 0). I don't know if my calculus professor would accept me "ignoring" sin and cos (he's very picky)

 

[mario99]

Hello, thank you for helping me. I was able to do the limit of the series using the confrontation test (the result was also 0). I don't know if my calculus professor would accept me "ignoring" sin and cos (he's very picky)

It is wrong to say the limit of the series. Dr.Peterson already mentioned that. Limit of the series means the sum of the series. You should say the limit of the sequence or the limit of divergence.

Hello, thank you for helping me. I was able to do the limit of the series using the confrontation test (the result was also 0). I don't know if my calculus professor would accept me "ignoring" sin and cos (he's very picky)

We can solve it without ignoring sin an cos.

The denominating term in the denominator is [imath]4^n[/imath], so divide every term in the numerator and the denominator by [imath]4^n[/imath].

[imath]\displaystyle \lim_{n\rightarrow \infty}\frac{2^n\sin^2(5n)}{4^n+\cos^2(n)} = \lim_{n\rightarrow \infty}\frac{\frac{2^n\sin^2(5n)}{4^n}}{\frac{4^n}{4^n}+\frac{\cos^2(n)}{4^n}} = \displaystyle \frac{\lim_{n\rightarrow \infty}\frac{2^n\sin^2(5n)}{4^n}}{1+0} = \lim_{n\rightarrow \infty}\frac{2^n\sin^2(5n)}{4^n} = \left(\lim_{n\rightarrow \infty}\frac{2^n}{4^n}\right) \left(\lim_{n\rightarrow \infty}\sin^2(5n)\right)[/imath]

Now you know that:

[imath]\displaystyle 0 \leq \left(\lim_{n\rightarrow \infty}\sin^2(5n)\right) \leq 1[/imath]

It is just a number, let us call it [imath]N[/imath], so we have:

[imath]0 \leq N \leq 1[/imath]

So our limit is:

[imath]\displaystyle \left(\lim_{n\rightarrow \infty}\frac{2^n}{4^n}\right) \left(\lim_{n\rightarrow \infty}\sin^2(5n)\right) = \left(\lim_{n\rightarrow \infty}\frac{2^n}{4^n}\right) \left(N\right) = Ne^{-\infty} = N*0 = 0[/imath]

 

[Dr.Peterson]

he also said we need to do a divergent test, which I didn't include because I didn't know how to calculate the limit of the series, any help with this too would be great

If your concern is how to show that the limit of the terms is zero, you already did that in your comparison:



Here you show that each term is less than the corresponding term of (1/2)^n; the squeeze theorem finishes the job.

This is part of what I meant in saying,

I see no need to separately do a divergence test, because comparison to a geometric series already incorporates that

Whenever you compare to a convergent series, what you are comparing to must go to zero, so your sequence must also!