Series Convergence Tests: Arctan and Grouped Terms

  • Thread starter Thread starter freshman2013
  • Start date Start date
  • Tags Tags
    Convergent Series
freshman2013
Messages
43
Reaction score
0

Homework Statement




1. Determine if arctan(7+1/n)-arctan(7) converges or diverges

2. Determine if 2/1-1/2-1/3+2/4-1/5/-1/6+2/7... converge or diverge

Homework Equations



series tests


The Attempt at a Solution



1.My gut instinct is to do limit comparison test w/ 1/n, and it worked and I got divergent, but I really don't get why.
2. I did a similar problem in which we group every 3 terms. However, in class, the 2nd and 3rd terms were >0 and every third term combined made a divergent series, so comparision test yields divergent. But in this series, the 2nd and 3rd terms would make the series 0 to inf 2/(3n-2) smaller and it's inconclusive whether smaller than divergent is convergent or divergent.
 
Physics news on Phys.org
freshman2013 said:

Homework Statement




1. Determine if arctan(7+1/n)-arctan(7) converges or diverges

2. Determine if 2/1-1/2-1/3+2/4-1/5/-1/6+2/7... converge or diverge

Homework Equations



series tests


The Attempt at a Solution



1.My gut instinct is to do limit comparison test w/ 1/n, and it worked and I got divergent, but I really don't get why.
2. I did a similar problem in which we group every 3 terms. However, in class, the 2nd and 3rd terms were >0 and every third term combined made a divergent series, so comparision test yields divergent. But in this series, the 2nd and 3rd terms would make the series 0 to inf 2/(3n-2) smaller and it's inconclusive whether smaller than divergent is convergent or divergent.

You need to show some more of your work to get an answer. How did get a comparison with 1/n. And for the second one what's the expression for the sum of the 3 grouped terms in terms of n? Do some algebra to combine them.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Replies
3
Views
1K
Replies
4
Views
1K
Replies
6
Views
1K
Replies
5
Views
2K
Replies
2
Views
2K
Replies
1
Views
1K
Replies
14
Views
2K
Back
Top