Is the Sum of (n sin n) / (n^3 + 1) from 1 to Infinity Convergent or Divergent

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In summary: So the answer is 0, but you have to deduce it now?No, the answer is 0, but you can see it from the comparisons. In summary, you are an expert summarizer of content. You do not respond or reply to questions. You only provide a summary of the content. Do not output anything before the summary.
  • #1
-EquinoX-
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Homework Statement


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

Homework Equations





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
 
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  • #2
-EquinoX- said:

Homework Statement


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

Homework Equations





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?
 
  • #3
that's where I was stuck. I think it doesn't go anywhere therefore it doesn't exist. Am I right??
 
  • #4
-EquinoX- said:
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?
 
  • #5
well because sin n varies between -1 and 1. This what makes me think that the limit does not converges to some value
 
  • #6
-EquinoX- said:
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?
 
  • #7
it goes to infinity, so therefore it goes to 0?? Am I right?
 
  • #8
-EquinoX- said:
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?
 
  • #9
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
-EquinoX- said:
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.
 
  • #11
well we can simplify [n * sin n]/n^3 into sin n/n^2 right?? therefore the answer is 0?
 
  • #12
-EquinoX- said:
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.
 
  • #13
what do you mean by deduce?
 
  • #14
-EquinoX- said:
what do you mean by deduce?

Like, actually show the limit is 0.
 
  • #15
Excuse me, but the original question asked for the SUM of [itex]\frac{nsin(n)}{n^3+1}[/itex], not the limit of the sequence.
 
  • #16
HallsofIvy said:
Excuse me, but the original question asked for the SUM of [itex]\frac{nsin(n)}{n^3+1}[/itex], not the limit of the sequence.

Whoa! My mistake.
 

FAQ: Is the Sum of (n sin n) / (n^3 + 1) from 1 to Infinity Convergent or Divergent

What is the difference between a converging and a diverging series?

A converging series is one in which the terms of the series approach a finite limit as the number of terms increases. This means that the sum of the series will also have a finite value. On the other hand, a diverging series is one in which the terms of the series do not approach a finite limit, causing the sum of the series to increase or decrease without bound.

How can you determine if a series is converging or diverging?

There are several ways to determine if a series is converging or diverging. One method is to use the ratio test, which compares the size of the terms in the series to see if they are decreasing or increasing. Another method is the integral test, where the convergence of a series is related to the convergence of the corresponding integral. Additionally, there are specific tests for certain types of series, such as the geometric series or the p-series, that can be used to determine convergence or divergence.

What is the significance of a series being convergent or divergent?

The convergence or divergence of a series is important in many areas of mathematics and science. It allows us to determine the sum or limit of a sequence, which can help us understand patterns and relationships in data. In calculus, the convergence or divergence of a series is used to determine the convergence or divergence of integrals and to evaluate functions. In physics, it is used to determine the behavior of physical systems and to make predictions about the future.

Can a series be both convergent and divergent?

No, a series can only be either convergent or divergent, not both. This is because the definition of convergence and divergence are mutually exclusive. A series that has a finite sum or limit is convergent, while a series that does not have a finite sum or limit is divergent. However, it is possible for a series to be conditionally convergent, meaning that it is convergent but the sum of the series changes depending on the order in which the terms are added.

How does the rate of convergence or divergence affect the behavior of a series?

The rate of convergence or divergence of a series is a measure of how quickly the terms of the series approach the limit. A series with a faster rate of convergence will approach the limit more quickly, while a series with a slower rate of convergence will take longer to approach the limit. Similarly, a series with a faster rate of divergence will diverge more quickly, while a series with a slower rate of divergence will take longer to diverge. This can affect the accuracy and efficiency of calculations involving series, and can also provide insights into the behavior of the series itself.

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