Convergent and Divergent Series

Date: 07/14/98 at 20:58:52
From: Gena
Subject: Convergent and Divergent Series

I've tried doing these problems and reading the lessons in the 
textbook, however I cannot seem to solve these problems.

Determine whether convergent or divergent
  1           1           1
-----    +  ------   +  ------   + ...
(1)(2)      (3)(4)      (5)(6)

Use the comparison test

1/2  + 2/3  + 3/4 + ...

Also if you could help me find more information on this topic I would 
really appreciate it.

Date: 07/15/98 at 10:36:37
From: Doctor Rob
Subject: Re: Convergent and Divergent Series

You can compare the first series, term by term, with

   1/1^2 + 1/2^2 + 1/3^2 + ...,

which, I hope you know, converges. The second series can be compared
term by term with

   1/2 + 1/3 + 1/4 + ...,

which, I hope you know, diverges.

With the first series, there is a trick.  1/(k*[k+1]) = 1/k - 1/(k+1),
so it can be rewritten

   1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ...

As an alternating series whose terms go to zero, it must converge to
something. It turns out that that something is ln(2).

As far as general advice, the comparison test can be used if

   lim a(n)*n > L > 0

to show divergence, by comparing with the harmonic series multiplied 
by L, as above. Similarly, it can be used if

   lim a(n)*n^[1+e] < L < infinity

with e > 0, to show convergence, by comparing with the series 
Sum L/n^(1+e), which converges if e > 0.

It cannot be used in conjunction with a series of the form 
a(n) = 1/n^(1+e) if lim a(n)*n = 0 but lim a(n)*n^(1+e) = infinity 
for any e > 0.  (Example: Sum 1/[n*log(n)].)  It might be used with 
some other series known to converge or diverge. It may or may not be 
used if the indicated limits do not exist.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/