Expert: Paul Klarreich - 5/4/2007

Question
Hi,
For waht values of a, if any, do the series converge?

segma from n=1 to infinity ( (1/(n-1) - (2a/(n+1) )

i'm using the integral test, i've integrated the series but i couldn't find the values that i should replace "a" with.
please help

Answer
Questioner:   Ibrahim
Category:  Calculus

 
Subject:  Infinite sequences and series
Question:  Hi,
For what values of a, if any, do the series converge?

segma from n=1 to infinity ( (1/(n-1) - (2a/(n+1) )

i'm using the integral test, i've integrated the series but i couldn't find the values that i should replace "a" with.
please help

I think you want a sum from  n = 2 to infinity, since  1/(n-1) cannot have  n=1.  That aside, you will probably want to do a little algebra:

 1      2a
----- - ----- =
n - 1   n + 1

n + 1 - 2a(n - 1)
-----------------  
n^2 - 1

n + 1 - 2an + 2a
-----------------  
n^2 - 1

n(1 - 2a) + 1 + 2a
------------------  
n^2 - 1

Separate into two terms again:

n(1 - 2a)     1 + 2a
--------- +  --------  
n^2 - 1      n^2 - 1

The second term, if you apply the integral test or a p-test, or whatever, will converge.  

But the first term behaves like the harmonic series and will diverge, unless  1- 2a = 0, in which case it is identically zero.

So I am afraid you will need  a = 1/2 for that.