Does the Limit of sin(n*alpha) as n Approaches Infinity Diverge?

  • Thread starter Thread starter AndrejN96
  • Start date Start date
  • Tags Tags
    Limit
Click For Summary
SUMMARY

The limit of sin(n*alpha) as n approaches infinity diverges for any fixed alpha in the interval (0, pi). The discussion clarifies that while specific values of alpha, such as pi/3, can be evaluated, the general case requires understanding that alpha is a constant within the specified range. It is emphasized that limits themselves do not diverge; rather, sequences can diverge, indicating that sin(n*alpha) does not approach a finite limit.

PREREQUISITES
  • Understanding of limits and sequences in calculus
  • Familiarity with trigonometric functions, specifically sine
  • Knowledge of the behavior of sin(n) as n approaches infinity
  • Concept of fixed versus variable parameters in mathematical expressions
NEXT STEPS
  • Study the properties of trigonometric limits in calculus
  • Explore the concept of divergence in sequences and series
  • Learn about the implications of fixed versus variable parameters in mathematical functions
  • Investigate the behavior of sin(n*alpha) for various fixed values of alpha
USEFUL FOR

Students studying calculus, particularly those focusing on limits and sequences, as well as educators seeking to clarify the distinction between limits and sequence behavior.

AndrejN96
Messages
26
Reaction score
0

Homework Statement


Show that lim n->inf sin(n*alpha), 0 < alpha < pi, diverges.

Homework Equations


lim n-> inf sin(n) diverges

The Attempt at a Solution


I know how to solve this for a constant value of alpha (i.e pi/3), but am unaware of solving this one, where alpha may vary. I know that sin(alpha) when alpha is in the given interval has a positive value, but don't know how to apply that to this solution.
 
Physics news on Phys.org
If you can do it for specific values of alpha, you should be able to use the same method for any value of alpha (you say "am unaware of solving this one, where alpha may vary"- alpha is NOT varying in a specific sequence- it can be any number but is a fixed number).

By the way- you want to show that the sequence diverges, that it has NO limit, not that "the limit diverges".
 
  • Like
Likes   Reactions: theodoros.mihos
AndrejN96 said:

Homework Statement


Show that lim n->inf sin(n*alpha), 0 < alpha < pi, diverges.

Homework Equations


lim n-> inf sin(n) diverges

The Attempt at a Solution


I know how to solve this for a constant value of alpha (i.e pi/3), but am unaware of solving this one, where alpha may vary. I know that sin(alpha) when alpha is in the given interval has a positive value, but don't know how to apply that to this solution.

As HallsofIvy has indicated, limits do not "converge" or "diverge"; they just either exist or do not exist. However, sequences can converge or diverge.
 

Similar threads

Prove that the limit as n->infinity of n^n/n! is infinity
  • · Replies 7 ·
Replies
7
Views
2K
The integral of (sin x + arctan x)/x^2 diverges over (0,∞)
  • · Replies 5 ·
Replies
5
Views
3K
Why does (-1)^n(sin(pi/n)) converge when (sin(p/n)) diverges
  • · Replies 2 ·
Replies
2
Views
10K
How to Approach Finding the Limit of an Equation as n Approaches Infinity?
  • · Replies 3 ·
Replies
3
Views
2K
Why Does the Series ∑ tan(1/n) Diverge?
  • · Replies 4 ·
Replies
4
Views
20K
Limit Definition of Derivative as n Approaches Infinity
  • · Replies 20 ·
Replies
20
Views
3K
Limit of Taylor Polynomial for Tn(x) as n Approaches Infinity
  • · Replies 2 ·
Replies
2
Views
1K
Can the limit of a quotient of trig functions approach a specific value?
  • · Replies 2 ·
Replies
2
Views
2K
Convergence of oscillatory/geometric series
  • · Replies 1 ·
Replies
1
Views
1K
Why Does the Limit of (n! / n^2) Approach Infinity?
  • · Replies 6 ·
Replies
6
Views
3K