Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Test for Convergence


Date: 8/20/96 at 17:28:15
From: Anonymous
Subject: Test for Convergence

I cannot solve this problem:

Test the following series for convergence.

sum {from k=0 to infinity} [{log(k+1)-log k}/tan^(-1) (2/k)]

Tan^(-1) is tangent inverse. Because -1 < tan^(-1) < 1, I used the 
comparison test, but I did not achieve the goal.

Thanks for your kindness.


Date: 8/21/96 at 15:49:36
From: Doctor Anthony
Subject: Test for Convergence

This series appears to diverge since the term when k -> infinity is 
not zero.

When k->inf  term is ln(1+1/k)/tan^(-1)(2/k)

As this tends to 0/0 we can use l'Hopital's rule to find its limiting 
value.

Differentiating top and bottom separately we get: 


   {1/(1+1/k)}(-1/k^2)
   -------------------
  {1/(1+4/k^2)}(-2/k^2)

 = (1/2){k/(k+1)}
   --------------
    k^2/(k^2+4)  

 =  (1/2)k(k^2+4)
    -------------
     k^2(k+1)

 = (1/2)(k^2+4)
   ------------
      k^2 + k

 = (1/2)(1+4/k^2)
   --------------
      1 + 1/k

If we let k->inf. This ratio -> 1/2

We can only get convergence if the terms tend to zero as k-> infinity, 
and this is not happening, so the series diverges.

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Sequences, Series

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/