# Stuck with a limit comparison test problem.

• icesalmon
In summary, the homework statement is trying to determine whether the series sin(1/n), sin(1/2), sin(1/3)... oscillates or not. If it does not oscillate, then the limit comparison test says it should be convergent, but if it does oscillate, then the limit comparison test says it should be divergent.
icesalmon

## Homework Statement

Determine the Convergence or Divergence of $$\sum_{n=1}^\infty\left(sin(1/n)\right)$$

## Homework Equations

limit comparison test

## The Attempt at a Solution

I don't know what to compare this series to, I tried the harmonic series to get sin(1/n)/(1/n) = nsin(1/n) which is just an infinity that oscilates. Which I believe to be divergent.
I've been really confused about a lot of comparisons because I'm only aware of a handful of series that converge or diverge the harmonic series (1/n) another series (1/n^2) which converges to pi2/6 p: series which if p > 1 it converges if 0 < p < 1 the p series diverges, geometric series which converge according to the formula a1 / (1 - r) if 0 < |r| < 1 and diverges if |r| > 1 And the Telescoping Series which converge according to the terms that cancel out in the set of real numbers it generates.

sin(1/n) doesn't really oscillate, does it? I think it approaches zero monotonically. You've got the right limit comparison there. sin(x)/x approaches 1 as x->0, doesn't it?

it's not monotonic because of the fact that it does oscillate, even if it does approach zero. lim[x -> 0] sin(x)/x = 0 by squeeze theorem, yes. My question is can I let u = 1/n to get sin(u)/u to let that happen? that would diverge because 0 is not positive. I'm also worried there would be a problem about the fact that I substituted in the first place.

Last edited:
icesalmon said:
it's not monotonic because of the fact that it does oscillate, even if it does approach zero. lim[x -> 0] sin(x)/x = 0 by squeeze theorem, yes. My question is can I let u = 1/n to get sin(u)/u to let that happen? that would diverge because 0 is not positive. I'm also worried there would be a problem about the fact that I substituted in the first place.

Um, the series sin(1/1), sin(1/2), sin(1/3)... does NOT oscillate. Other series involving sin do oscillate, but this isn't one of them. And sin(x)/x doesn't approach 0 as x->0.

-1 < sin(x) < 1
-1/x < sin(x)/x < 1/x
as x approaches infinity it does equal zero by squeeze theorem, as x approaches zero it approaches 1, and I would be taking the limit as x approaches infinity, not zero. Isn't that what I do with limit comparison tests, it's at least what I've been doing with the other ones.

icesalmon said:
-1 < sin(x) < 1
-1/x < sin(x)/x < 1/x
as x approaches infinity it does equal zero by squeeze theorem, as x approaches zero it approaches 1, and I would be taking the limit as x approaches infinity, not zero. Isn't that what I do with limit comparison tests, it's at least what I've been doing with the other ones.

1/n doesn't approach infinity as n->infinity. It approaches 0! That's the limit you should be caring about, isn't it??

nono you're right, i agree, the answer says it should be convergent though. I keep getting divergent...is this book wrong?

icesalmon said:
nono you're right, i agree, the answer says it should be convergent though. I keep getting divergent...is this book wrong?

If the books says it's convergent, then the book is wrong.

thanks for helping me clear this up, but my takeaway from this would be to ask how to figure out, when using direct comparisons and limit comparisons, what to compare the series to. My professor asks the question "what does an look like?" that is a hard question to answer when I have something like ln(n)/(n+1) because the comparison has nothing to do with ln(n)

icesalmon said:
thanks for helping me clear this up, but my takeaway from this would be to ask how to figure out, when using direct comparisons and limit comparisons, what to compare the series to. My professor asks the question "what does an look like?" that is a hard question to answer when I have something like ln(n)/(n+1) because the comparison has nothing to do with ln(n)

I'd say ln(n)/(n+1) 'looks like' ln(n)/n. Because the '+1' is unimportant. But I would still compare it with 1/(n+1) because I know ln(n)>1 for n>e and I know 1/(n+1) diverges. You could also compare it with ln(n)/(n+n) and show that diverges by an integral test. There's more than one way to do these problems.

I'll keep that in mind, and come back for questions. Thanks, for your help.

## What is a limit comparison test and how does it work?

A limit comparison test is a method used in calculus to determine the convergence or divergence of a series. It compares the ratio of the terms in a given series to the terms in a known convergent or divergent series. The limit of this ratio is then taken to determine the convergence or divergence of the original series.

## When should I use a limit comparison test?

A limit comparison test should be used when the series you are trying to determine the convergence or divergence of is complex and does not fit into one of the standard tests (such as the comparison test or the integral test). It is also useful when the series contains both positive and negative terms.

## What are the steps to apply a limit comparison test?

The steps to apply a limit comparison test are as follows:
1. Determine the given series and the known series to be compared with.
2. Take the ratio of the terms in the given series to the terms in the known series.
3. Find the limit of this ratio as n approaches infinity.
4. If the limit is a positive finite number, then the series converges. If the limit is 0 or infinity, then the series diverges. If the limit is inconclusive, then the test is inconclusive and another method should be used.

## What are some common mistakes when using a limit comparison test?

Some common mistakes when using a limit comparison test include:
- Forgetting to take the absolute value of the terms before finding the ratio
- Using the wrong known series for comparison
- Not simplifying the ratio before taking the limit
- Applying the test to a series that does not meet the conditions for convergence or divergence
- Using the test as the only method for determining convergence or divergence, without considering other tests

## Can I use a limit comparison test for alternating series?

Yes, a limit comparison test can be used for alternating series. However, you must first convert the alternating series into an equivalent non-alternating series by taking the absolute value of the terms. Then, follow the steps for applying the limit comparison test as usual.

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