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### 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/
```
Associated Topics:
High School Analysis
High School Sequences, Series

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