• Support PF! Buy your school textbooks, materials and every day products Here!

Limit of sin(2x) and ln(x)

  • Thread starter Mikabird
  • Start date
  • #1
2
0

Homework Statement



[tex]\Sigma[/tex] sin(2n)/(n*ln(n)^2) from n=2 to [tex]\infty[/tex] Determine if the summation converges or diverges.

Homework Equations



I cannot even figure out where to start. Should I use the Comparison Test?

The Attempt at a Solution



I thought that I could compare it to sin(2x)/2x but when n=2, sin(2x)/2x is too small.
Where am I messing up?
 
Last edited:

Answers and Replies

  • #2
867
0
The comparison test among others only tests whether the sum converges. If you have to find the limit of the sequence, you need to try something else, probably with L'hopital's rule.
 
  • #3
2
0
oops, I meant find whether the summation converges or diverges. Thanks.
 
  • #4
867
0
Then I think the easiest would be finding another series with which to compare the original one and then use the integral test.
 
  • #5
33,158
4,842
The comparison test among others only tests whether the sum converges.
This is not true. The comparison test can be used also to test a series that diverges.

The key to using the comparison test is to have a good idea first whether a series converges or diverges.

In the following, I'm assuming that you're dealing with a series of nonnegative terms. If you are reasonably sure that [itex]\sum a_n[/itex] converges, and cn is the general term in a convergent series, and an <= cn, then your series converges.

On the other hand, if you believe that [itex]\sum a_n[/itex] diverges, and dn is the general term in a divergent series, and an >= dn, then your series diverges.
 
  • #6
867
0
You're right, and I've no excuse for not remembering to include that they test for divergence. :redface:
 
  • #7
1,838
7
This is easy, because the series is absolutely convergent. By the triangle inequality, the absolute value of a summation is equal to or less than the summation of the absolute values. Then you can use that |sin(2n)| < 1. Since the summation of 1/[n ln^2(n)] converges (e.g. use the integral test), you find that the summation converges.
 

Related Threads for: Limit of sin(2x) and ln(x)

Replies
1
Views
4K
  • Last Post
Replies
3
Views
660
Replies
4
Views
12K
Replies
8
Views
26K
Replies
10
Views
4K
Replies
13
Views
23K
  • Last Post
Replies
14
Views
4K
Replies
9
Views
56K
Top