Convergence of sin(1/n) Summation from n=1 to Infinity

  • Thread starter Frillth
  • Start date
  • Tags
    Convergence
In summary, the conversation discusses finding the convergence of the sum of sin(1/n) from n=1 to infinity. The person is struggling to prove it and is considering using the limit comparison test or comparison test. They attempt the limit comparison test with 1/n and conclude that sin(1/n) must diverge. However, further clarification is needed on how they arrived at this conclusion.
  • #1
Frillth
80
0
Okay, this is my last problem today. I swear.

Homework Statement



For this problem, I need to find the convergence of the sum of sin(1/n) from n=1 to infinity.

Homework Equations



None

The Attempt at a Solution



I know that this has to converge, but I'm having a hard time proving it. It seems like I should either be using the limit comparison test or just the comparison test, but I can't think of whta I can compare it to.
 
Physics news on Phys.org
  • #2
This is very counter-intuitive. I just tried the limit comparison test with 1/n as follows:

lim sin(1/n) = cos(1/n) * -1/n^2 = cos(1/n) = 1
n->inf 1/n = -1/n^2

That means that sin(1/n) must diverge. Is this right?
 
  • #3
I don't follow anything from your second post, except the conclusion, which is correct. But please describe more clearly how you arrived at it.
 

FAQ: Convergence of sin(1/n) Summation from n=1 to Infinity

What is the definition of "Convergence of sin(1/n)"?

The convergence of sin(1/n) refers to the behavior of the sine function when the input values approach infinity, or when the input values get closer and closer to zero.

Does sin(1/n) converge or diverge?

The function sin(1/n) converges as n approaches infinity. This means that the output values of the function approach a finite limit as n gets larger and larger.

What is the limit of sin(1/n) as n approaches infinity?

The limit of sin(1/n) as n approaches infinity is 0. This means that the output values of the function get closer and closer to 0 as the input values get larger and larger.

How is the convergence of sin(1/n) related to the convergence of 1/n?

Since sin(1/n) is the sine function of 1/n, the convergence of sin(1/n) is directly related to the convergence of 1/n. As 1/n approaches infinity, sin(1/n) also approaches infinity.

What are some real-world applications of the convergence of sin(1/n)?

The convergence of sin(1/n) has applications in fields such as physics, engineering, and economics. For example, it can be used to model the behavior of waves, the trajectory of projectiles, and the growth of populations.

Similar threads

Root test proof using Law of Algebra
Replies
6
Views
1K
Does the Series ## \frac {(n^3+3n)^{1/2}} {5n^3+3n^2+2 sin (n)}## Converge?
Replies
29
Views
4K
Proving convergence of rational sequence
Replies
2
Views
915
Why does (-1)^n(sin(pi/n)) converge when (sin(p/n)) diverges
Replies
2
Views
9K
Prove that the limit as n->infinity of n^n/n! is infinity
Replies
7
Views
607
Study the convergence and absolute convergence of the following series
Replies
5
Views
1K
Finding the r value for the series to converge
Replies
15
Views
2K
Limit of probabilities of a large sample
Replies
1
Views
696
Special Comparison Test For Infinite Series
Replies
4
Views
782
Convergence: Epsilon-N Definition
Replies
3
Views
2K

Hot Threads

Recent Insights

Back
Top