Using the Comparison Test to Determine if a Series Converges

Date: 05/19/2005 at 00:14:36
From: Susan
Subject: Does this series converge?

Determine, with proof, whether or not the following series converges:

   sum n = 3 to infinity of 1/[[ln(n)]^ln(n)]

I tried the Integral Test, Test for Divergence, and Ratio Test, but 
they seem to not work or not be able to prove anything.  I'm 
considering the Comparison Test but I don't know what to use for a
comparison.

Date: 05/19/2005 at 20:31:33
From: Doctor Vogler
Subject: Re: Does this series converge?

Hi Susan,

Thanks for writing to Dr. Math.  That's a good question.  The 
Comparison Test is a very powerful test but is usually saved for a
last resort for one main reason: there aren't very many good methods
for deciding what to compare your series to.

So I will give you one method.  But first I would like to mention a
different method: many people try to compare their series to a series
that has a very similar form but is slightly more simple.  For
example, they might compare

   n^2 + 2n + 5
 ----------------
  n^5 - 6n^2 + 7

to

   2n^2
  ------.
    n^5

This is a fine idea, especially since showing that one series is
bigger or smaller than another is relatively simple when they have a
similar form.  But there is a problem with this method: sometimes the
form doesn't generalize well.  Sometimes making the series more simple
requires changing the form quite a bit.  In those cases, we need a new
method.

So here is a new method.  Perhaps I could show that my series is
smaller than

  1/n^5,

and this would show that my series converges.  Great!  But if my
series is smaller than 1/n^5, then it is also smaller than

  1/n^2,

because 1/n^2 is bigger than 1/n^5.  So all I really need to do is
decide what is the biggest series that still converges.  Then I can
compare any series to this.  And I could try to find the smallest
series that diverges.  Then I can compare any series to this.

In fact, there isn't a "largest" convergent series, nor a "smallest"
divergent series, but we can get pretty close.  The harder question is
the first one, since we can always multiply a convergent series by
some big number and get a larger sum, but in some sense this isn't
what we mean.  For example, the series 1/n^2 sums to pi^2/6 or about
1.645, while 10*(1/2)^n sums to 10 exactly.  But 1/n^2 is always
larger than 10*(1/2)^n when n is large.  The Comparison Test is only
concerned with large n, so really 1/n^2 is the larger series.

But let's think about a "small" divergent series.  That means that the
partial sums go to infinity.  But we know some functions that go to
infinity quickly (like n^2 and e^n) and others that go to infinity
much more slowly (like ln n and ln ln n).  What series have partial
sums that go to infinity very slowly?

So some diverging series that diverge very slowly are

  1/n

whose partial sums are about ln n, and

     1
  -------
  n(ln n)

whose partial sums are about ln ln n, and

          1
  ----------------
  n(ln n)(ln ln n)

whose partial sums are about ln ln ln n, and so on.  If you think that
your series is divergent, then it will almost certainly be bigger than
one of these series.

For large convergent series, we could just take something slightly
smaller than the above series.  For example, slightly smaller than

  1/n

is

     1
  -------
  n^(1+e)

for any positive number e.  But in most cases, it is just fine to take
the most simple one, with e=1, namely

  1/n^2.

When this is already too small, then we can go another step and use a
series slightly smaller than

     1
  -------
  n(ln n)

such as

      1
  ---------.
  n(ln n)^2

And if this is still too small, then we can use

          1
  ------------------
  n(ln n)(ln ln n)^2

or

                1
  ------------------------------.
  n(ln n)(ln ln n)(ln ln ln n)^2

In most cases, these will be more than sufficient.

So let's summarize.  If we have a series and we're not quite sure how
fast it grows, and whether it converges, then we should check if (for
sufficiently large n) it is smaller than

  1/n^2

or larger than

  1/n.

If so, then we are done.  If it is between these two, then we go 
another step and check if our series is smaller than

     1
  -------
  n(ln n)

or larger than

      1
  ---------.
  n(ln n)^2

If so, then we are done.  If not, then we keep going.


Finally, we have only one big question to answer.  How do we tell if
our series is bigger or smaller than another?  Well, some of this has
to come by experience, but I'll give you a few pointers.  First of
all, remember that the Comparison Test only requires the inequality to
hold for "sufficiently large n," that is, for n bigger than some
finite number.  So when you are doing the comparison, think of n as
being very very large.  Secondly, especially when there are exponents
involved, logarithms are very useful.  Logarithms are your friend. 
Remember that the log function, and its inverse the exp (or e^x)
function, are both increasing functions, which means that they
preserve the inequality.  That is,

  a > b  if and only if  ln a > ln b.

Use this to your advantage.  Now let's do your function.  In fact, it
turns out that your function is quite small, because

  (ln n)^(ln n)

is bigger (for sufficiently large n) than any polynomial.  How do you
tell?  Well, we check:

  Is  1/[[ln(n)]^ln(n)] < 1/n^2  ?

We can either cross-multiply and find that this is the same as

  n^2 < (ln n)^(ln n)

and then take logarithms, or we can take logarithms right away

  - ln [(ln n)^(ln n)] < - ln (n^2)

which is equivalent to

  ln (n^2) < ln [(ln n)^(ln n)]

  2 ln n < (ln n)(ln(ln n))

  2 < ln ln n

  e^(e^2) < n

and we find that when n is large (bigger than e^(e^2)), then this
inequality is always satisfied, and therefore the first one

  1/[[ln(n)]^ln(n)] < 1/n^2

is as well.  Therefore, the Comparison Test tells us that your series
converges.

I hope that you have found this discussion helpful, and I hope it
allows you to solve similar problems in the future.  If you have any
questions about this or need more help, please write back and show me
what you have been able to do, and I will try to offer further
suggestions.

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

Date: 05/19/2005 at 22:23:45
From: Susan
Subject: Does this series converge?

Is there any way that I can apply the Limit Comparison Test to this 
problem (where the function you choose to compare with does not have 
to be greater than the original and just take the limit going to 
infinite of Bn/An (Bn being the comparing function and An being the 
original)and if the limit comes out to anything between 0 and 
infinite than An behaves (converges or diverges) like Bn)?

Date: 05/20/2005 at 17:23:10
From: Doctor Vogler
Subject: Re: Does this series converge?

Hi Susan,

The Limit Comparison Test is used to show that your series grows at
exactly the same rate as some other series.  In other words, you can't
just use a sufficiently large convergent series, or a sufficiently
small divergent series; rather you need a series that grows at exactly
the same rate.

For something like your series, the rate of growth of the series is

  (ln n)^(-ln n),

and you can't make this any simpler and still preserve the same rate
of growth.

In other words, no, I can't see any way to apply the Limit Comparison
Test to your series.  But once you confirm that your series converges,
then you could use the Limit Comparison Test to determine that various
similar series also converge.

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