# Prove if sin(1/x) diverges or converges

I want to prove if sin(1/x) diverges or converges as x approaches infinity. I know the answer is diverge because the function oscillates. I want to prove it more rigorously though, for my own well being and for the exam coming up. I am having trouble with the ratio test and I rewrote the expression as
sin(1/x)[sin^2(1/x) +cos^2(1/x)] to see if it would help but that won't because if I set b equal to sin^2(1/x) +cos^2(1/x) that is not smaller than the original expression of course so I can't use the direct comparison test. I assume I will be using the limit comparison test but I need a little kick start. Thanks.

jbunniii
Homework Helper
Gold Member
I want to prove if sin(1/x) diverges or converges as x approaches infinity. I know the answer is diverge because the function oscillates.
Are you sure about that? What happens to ##1/x## as ##x \rightarrow \infty##?

jbunniii
Homework Helper
Gold Member
I assume I will be using the limit comparison test but I need a little kick start. Thanks.
Do you know this inequality? ##|\sin(x)| \leq |x|##, valid for all ##x##.

Sin(x) is a bounded function with an upper bound of 1 and a lower bound of -1. As values of x take values close to 0 then sin(x) takes values also close to 0. So in your case when x takes really big values close to infinity what does 1/x take values close to and what does that say for sin(1/x).

Both of you are saying exactly what I thought at first. I thought the function would converge because the limit of 1/x as x approaches infinity is zero. Sine of zero is zero. So I concluded that the sum converges. When I looked at the answer in the back of the book it said the series was divergent. I believed this to be true against my initial intuition when I looked at a graph of the function it does oscillate as x approaches infinity, even when I made the viewing window quite large.
jbunniii, yes I do know that inequality. All these things are the reason I don't see the answer. I need to see clarity in this problem.

D H
Staff Emeritus
What series are you talking about?

The original post only asks about sin(1/x) as x tends to infinity. That has a well defined limit. If you are asking about something such as ##\sum_{n=1}^{\infty} \sin(\frac 1 n)##, that's a very different question than ##\lim_{x\to \infty} \sin(\frac 1 x)##.

I'm sorry. Yes i'm talking about the sum you have written above from n=1 to infinity.

D H
Staff Emeritus
You should have said so in your original post! Be specific when you ask a question. Use the proper notation. We can't read your mind.

Just because sin(1/x) converges to zero as x→∞ (which it does) does not mean that Ʃ sin(1/n) converges (which it doesn't).

Dick
Homework Helper
I'm sorry. Yes i'm talking about the sum you have written above from n=1 to infinity.

If x is very close to zero (like 1/n is for n large) then sin(x) is very close to being x. Look at the graphs near 0. Does that suggest a way to create a comparison series that is smaller but diverges?

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The graph near zero is very intense. This is because the function will experiences infinitely many oscillations as it gets closer to zero correct? This problem is confusing. I can't use the alternating series test so my only option would be comparison like you said, correct? So I want a function that will be undefined at zero and most likely involves a trig function that is less than sin(1/n) that I can show the limit as n approaches infinity of this new series diverges. Do I compare it to another sine function? Since it's not an alternating series I know can't use sin(1/(n+1)). Plus, this series converges.

Dick
Homework Helper
The graph near zero is very intense. This is because the function will experiences infinitely many oscillations as it gets closer to zero correct? This problem is confusing. I can't use the alternating series test so my only option would be comparison like you said, correct? So I want a function that will be undefined at zero and most likely involves a trig function that is less than sin(1/n) that I can show the limit as n approaches infinity of this new series diverges. Do I compare it to another sine function? Since it's not an alternating series I know can't use sin(1/(n+1)). Plus, this series converges.

1/n is near 0 for large n. That's where you are concerned with an estimate of sin. I meant look at the graphs of sin(x) and x near 0.

Sorry it took so long to reply. I believe I have the answer though. If I use the limit comparison test with 1/n as my series to compare I get the correct answer. I didn't know that you could use expressions within a trigonometric function as your series to compare.

Dick