Need help with a (apparently) difficult series

  • Thread starter Thread starter Sebacide
  • Start date Start date
  • Tags Tags
    Series
Click For Summary
SUMMARY

The discussion focuses on the convergence of the infinite series $$\sum_{n=1}^{+\infty}\sin(n)\sin\left(\frac{1}{n}\right)\left(\cos\left(\frac{1}{\sqrt{n}}\right)-1\right)$$. Participants emphasize the necessity of applying bounds on ##\sin(n)## and utilizing small angle approximations for large ##n##, specifically ##\sin(1/n) \approx 1/n## and ##\cos(1/\sqrt{n}) \approx 1 - \frac{1}{2n}##. These approximations stem from Taylor series expansions, although concerns are raised regarding the speed at which terms converge to their approximated values in the context of infinite series.

PREREQUISITES
  • Understanding of Taylor series expansions
  • Knowledge of trigonometric functions and their limits
  • Familiarity with convergence tests for infinite series
  • Basic calculus, particularly in handling small angle approximations
NEXT STEPS
  • Study Taylor series and their applications in approximating functions
  • Learn about convergence tests for infinite series, such as the Ratio Test and Root Test
  • Explore the properties of trigonometric functions in calculus
  • Investigate the implications of using approximations in series convergence
USEFUL FOR

Mathematicians, calculus students, and anyone interested in the convergence of infinite series and the application of Taylor series in analysis.

Sebacide
Messages
3
Reaction score
0
Homework Statement
I have a series, but i cannot find a method to study the convergence. Can anyone, please, help me with this series? I can't understand what kind of method can be used to study its convergence.
Relevant Equations
I thought about using ##0\leq\left|\sin(n)\right|\leq1##
This is the series: $$\sum_{n=1}^{+\infty}\sin(n)\sin\left(\frac{1}{n}\right)\left(\cos\left(\frac{1}{\sqrt{n}}\right)-1\right)$$
 
Physics news on Phys.org
Good start. I'm sure that you will need to use the bound on ##\sin(n)## that you mention, but you need more than that. I think it must have something to do with the small angle approximations. For large ##n##, ##\sin(1/n) \approx 1/n## and ##\cos(1/\sqrt{n}) \approx 1- \frac{1}{2n}##. These approximations can be derived from the Taylor series expansions. But I have concerns about using approximations in an infinite series since I do not know how fast the terms approach their approximate value.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K