# How do I show sin 1 + sin 2 + sin 3 + diverges?

1. Mar 24, 2009

### incubusfan723

1. The problem statement, all variables and given/known data

I was given the series sin 1 + sin 2 + sin 3 + sin 4 + ... and I must show that it diverges. Can anyone point me in the right direction as to how to go about doing this?

2. Relevant equations

3. The attempt at a solution

2. Mar 24, 2009

### xcvxcvvc

If the parts of a series never reach zero, wouldn't you keep adding chunks, so the sum never settles down(never converges - it diverges)?

3. Mar 24, 2009

### incubusfan723

I understand it intuitively that the sum is never going to settle down due to the nature of the function, but i'm having a hard time putting that concept into a formal proof.

4. Mar 24, 2009

### Staff: Mentor

Note that the range of sin(x) is -1 to 1. So if you picked the arguments to the sin(x) function correctly, you could get the sum to not diverge.

The hard part of this qestion is that the x values are not an even fraction of PI...

5. Mar 24, 2009

### chroot

Staff Emeritus
6. Mar 24, 2009

### Staff: Mentor

7. Mar 24, 2009

### Staff: Mentor

Or more accurately, the first rule is talking about converging, as opposed to not diverging, which the OP is asking about.

8. Mar 24, 2009

### chroot

Staff Emeritus
No, berkeman. The summand in this case is sin(n), and the limit of sin(n) as n goes to infinity is undefined. Therefore, the series does not converge.

- Warren

9. Mar 24, 2009

### Staff: Mentor

I totally agree it does not converge. But the OP said the problem was to show that it diverges. The sum can oscillate without diverging.

I just did a quick Excel check to get a feel for what the sum does. After 100 terms, it's not obvious that it will diverge....

10. Mar 24, 2009

### xcvxcvvc

this is the right answer. to show it, take the limit of sin(x) as x approaches infinity (which is undefined)

the only way an infinite series of sine could converge is if each piece equaled zero. for example, the series of sin(k) when k = pi and k increased by 2pi each time. (0 + 0 + 0.... adds up to 0)

11. Mar 24, 2009

### Staff: Mentor

What they refer to in the wiki article as "Limit of the summand" is similar to one that appears in some calculus texts, the Nth Term Test for Divergence. I.e., in a series $\sum a_n$, if lim an is not zero, the series diverges.

A common mistake that students make happens when they find that lim an = 0, and conclude that $\sum a_n$ converges.

12. Mar 24, 2009

### Staff: Mentor

After 500 terms, the running series sum is spending more time postive than negative, but the highest it's getting is just under 2.

13. Mar 24, 2009

### xcvxcvvc

yeah, lol. i remember people confused about the topic. i just giggled inside.

if i am mad, my face is red.

i am not mad, so my face must not be red :D

wait, what about when a cute girl walks my way

14. Mar 24, 2009

### robphy

Can an Euler-type sum help?

S_n=sin 1 + sin 2 + sin 3 + sin 4 + ... +sin n
S_n=sin n + sin (n-1) + sin (n-2) + sin (n-3) + ... +sin 1

2S_n= (sin 1 + sin n) + (sin 2 + sin (n-1) ) +... + (sin n + sin 1)

Then, use sin a+sin b identities to get a closed form expression.
Study n -> infty.

15. Mar 24, 2009

### rwisz

I would just do what's already been mentioned and write the series as

$$\sum_{n=1}^{\infty}sin(n)$$

then just perform the nth test for divergence to show that the limit as n approaches infinity of sin(n) is infinity, and therefore NOT zero, and therefore it DIVERGES. And as otherwise mentioned, intuitively you know that sine from 1 to infinity will always oscillate from -1 to 1, and thus never converge to a single value.

Unless you have to write the proofs for the nth test for divergence... I'm not really sure what else there is to be done here...

16. Mar 24, 2009

### chroot

Staff Emeritus
At least in the mathematical sense, oscillation is a form of divergence, because the series sum never settles down to any specific value. The nth term test is all that's needed to show that this series diverges. I've even heard this called the "hurdle test" by math teachers, because if it doesn't pass this hurdle, it cannot converge, and there's no need to check anything else.

- Warren

17. Mar 25, 2009

### Staff: Mentor

I agree completely. A series doesn't have to have sums that become ever larger or ever more negative to diverge.

18. Mar 25, 2009

### MeTh0Dz

19. Mar 25, 2009

### Count Iblis

You can also evaluate:

$$S(N)=\sum_{n=0}^{N}\sin(n) = \text{Im}\sum_{n=0}^{N}\exp(i n)=\text{Im}\frac{1-\exp[i(N+1)]}{1-\exp(i)}$$

So,

S(N) = [sin(1) + sin(N) - sin(N+1)]/[2 - 2 cos(1)]

and we see that lim N to infinity of S(N) does not exist.

Last edited: Mar 25, 2009