# Coverging or diverging series

1. Mar 5, 2007

### -EquinoX-

1. The problem statement, all variables and given/known data
the sum of (n sin n)/(n^3+1)
n from 1 to infinity

2. Relevant equations

3. The attempt at a solution
what I have done so far is using the comparison method by comparing this with n sin n/ n^3, but I don't know what's the next step and even if my comparison is correct or not

2. Mar 5, 2007

### JasonRox

Have you heard of L'Hopital's Rule?

The comparing method your using is also another way.

So, what happens to [n*sin(n)]/n^3 when n goes to infinity?

3. Mar 5, 2007

### -EquinoX-

that's where I was stuck. I think it doesn't go anywhere therefore it doesn't exist. Am I right??

4. Mar 5, 2007

### JasonRox

What's your argument for it not going anywhere?

5. Mar 5, 2007

### -EquinoX-

well because sin n varies between -1 and 1. This what makes me think that the limit does not converges to some value

6. Mar 5, 2007

### JasonRox

Yes, but where does the denominator go?

7. Mar 5, 2007

### -EquinoX-

it goes to infinity, so therefore it goes to 0?? Am I right?

8. Mar 5, 2007

### JasonRox

Not so quick.

So, the denominator goes to infinity, and the numerator bounces around -1 and 1.

Can you replace sin(n) by something else? Like say, -1 or 1?

9. Mar 5, 2007

### -EquinoX-

yeah If I replace it with either -1 or 1 both will most likely to reach 0 from both sides, from the negative sides and the positive sides.

10. Mar 5, 2007

### JasonRox

So, why not just pick one?

So, we have 1/n^2, and [n*sin(n)]/n^3, and the original one.

Write down the comparisons, and you should be good to go.

11. Mar 5, 2007

### -EquinoX-

well we can simplify [n * sin n]/n^3 into sin n/n^2 right?? therefore the answer is 0?

12. Mar 5, 2007

### JasonRox

Yeah, the answer is 0, but you have to deduce it now. Or I think you do. Sometimes they just ask for the solution.

13. Mar 5, 2007

### -EquinoX-

what do you mean by deduce?

14. Mar 5, 2007

### JasonRox

Like, actually show the limit is 0.

15. Mar 6, 2007

### HallsofIvy

Excuse me, but the original question asked for the SUM of $\frac{nsin(n)}{n^3+1}$, not the limit of the sequence.

16. Mar 6, 2007

### JasonRox

Whoa! My mistake.