# Coverging or diverging series

• -EquinoX-

## Homework Statement

the sum of (n sin n)/(n^3+1)
n from 1 to infinity

## 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

## Homework Statement

the sum of (n sin n)/(n^3+1)
n from 1 to infinity

## 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

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?

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

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

What's your argument for it not going anywhere?

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

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

Yes, but where does the denominator go?

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

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

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?

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.

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.

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.

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

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

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

what do you mean by deduce?

what do you mean by deduce?

Like, actually show the limit is 0.

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

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

Whoa! My mistake.