using only BASIC tests for Convergence (comparison, root, ratio,
integral test, etc.) show that the following three series are (or are
not) convergent.
1.)  2 / [n * (ln n)^3]
2.) [tan (1/n)]/n
3.) [2^(1/n)]-1

1. Consider the function f(x) = 1/[x * (ln x)^3]. Clearly f(x) is
decreasing. Integrate this from 2 to k, then apply the substitution u
= ln x. You arrive at INT[du/(u^3), u from ln 2 to ln k]. So, by the
integral test it is clear to see that this series converges (remember,
the integral of 1/x^p converges for p > 1, the harmonic series being
the most striking exception).

I'll write back with 2 and 3 if I find anything later.

- Entropyx

Hi again.

Sorry I took so long to reply, but I hadn't given the problems too
much thought. I believe that I have a solution for #2. In order to
show that the series {tan(1/n)/n} converges, we will use a comparison
test, i.e. we will show that, much like 1/n^3:

       0 < tan(1/n)/n < 1/n^2

The fact that tan(1/n)/n > 0 for positive n is self-evident. So we
must show that:

       tan(1/n)/n < 1/n^2 ... or:
       tan(1/n) < 1/n

Or, we can say, is tan(x) < x? The answer here is a resounding yes.
For all -1 < x < 1, the Maclaurin series of tan(x) gives x + x^3/3 +
(2x^5)/15 + ... . We have this condition on x whenever n > 1. Thus,
tan(1/n)/n converges.

- Entropyx

Ha. Sorry, this is what I get for not sleeping. Obviously, tan(x) > x,
and thus I am completely wrong. Ignore the previous message.

- Entropyx

Hi, entropyx/entropix-ga:

You are not completely wrong, your solution to part 2) is basically correct.

Yes, tan(1/x) > 1/x on x = 1,2,3,... , but:

tan(1/x) = sin(1/x)/cos(1/x)

and cosine is decreasing on (0,1], so:

tan(1/x) < sin(1/x) / cos(1)

You can therefore estimate:

0 < tan(1/x) = sin(1/x)/cos(1/x) < (1/x)/cos(1)

as x runs through the positive integers.

'Nuff said?

regards, mathtalk-ga

This answer for #1 is great.  Thank you.  For 2 and 3, my prof was
looking for the Limit Comparison test.  Using 1/(n^2) for number 2
{giving: n*tan(1/n)} and 1/n for number 3 {giving n*2^(1/n)-n}.  This
shows that 2 converges and 3 does not.

I'll be away from the e-mail and the web for the next two weeks.  SO
MY NEXT RESPONSE WILL TAKE A WHILE.  Is there some Google Referee who
could step in and say what comes next.  I appreciate all the effort,
and the answer for #1 is spot-on.

Thanks again,

Truman

Hi, truman-ga:

Entropyx-ga is not a Researcher and thus could not post the partial
answer to 1 or 2 for payment.  It often happens that questions receive
partial or full answers as "Comments" rather than as Answers.  I'm
sure that Entropyx-ga appreciates the positive feedback on the answer
to part 1, and this from my own experience is a strong motivation to
post helpful comments.

A couple of ideas about "what comes next".  From the nature of your
last comment it seems that you no longer need an "Answer" to this
post.  For that reason you may want to click the "Close" button on
your page for this thread, which will "Expire" the question.  No
further Answer or Comments will be allowed.

On the other hand if you wanted additional comments, you could let the
Question go.  Any reasonable Researcher would read all the foregoing
Comments and probably realize your Question has already been answered
to some extent, and that not much further improvement could be given
in an "Answer".  When a Question is allowed to expire at its "natural"
30-day limit, it remains open for Comments.

regards, mathtalk-ga

P.S.  It may be a bit confusing to tell who is a Researcher and who
isn't, esp. when you are ready good comments.  Researchers will stand
out as having clickable links for their names in the headings of
posts; clicking their takes you to a summary of Questions previously
answered by that Researcher and their ratings.

Ah, indeed. Thank you mathtalk for fixing the error.

- Entropix (forget about the y thing)