Expert: Paul Klarreich - 12/3/2006

Question
Hi Paul,

I was given a a series from 3 to infinity of 2/(n(ln(n))*sqrt(ln(ln(n)))) and i am clueless where to start.
I am not sure how to treat the natural logs.
thanks,
Mike

Answer
Questioner:  Mike
Category:  Calculus
 
Subject:  Infinite series
Question:  Hi Paul,

I was given a series from 3 to infinity of 2/(n(ln(n))*sqrt(ln(ln(n)))) and i am clueless where to start.
I am not sure how to treat the natural logs.
thanks,
Mike
.......................
Hi, Mike,

I am not sure what you are supposed to do with the series.  (ALWAYS, ALWAYS, ALWAYS, include the instructions with the example.  If it comes out of a book that has something like:

In 23 to 55, do the...... :

23.
24.
...
55.

then be sure to include the instructions in the 23-55 part.

SO, the only thing I can think of doing with this is to test for convergence.  I will not even try to compute the sum (assuming it exists).

So you need a test for convergence for:

inf            2
Sum ------------------------
n=3  n ln n * sqrt(ln ln n)

Offhand the expression looks like something you might see in an integration example, like:


{inf          2 dx
|   ----------------------
}3  x ln x * sqrt(ln ln x)

So how about using the Integral Test, which says that

inf
Sum f(n)
n=a

converges if and only if the improper integral:

{inf
|   f(x) dx
}a  

converges.  So we try to integrate:

{inf          2 dx
|   ----------------------
}3  x ln x * sqrt(ln ln x)

How about letting  u = ln ln x?

u = ln ln x
du     1   1
-- = ---- ---
dx   ln x  x

du     1
-- = ------
dx   x ln x

      dx
du = ------
    x ln x
And the integral becomes:

{   2 du
|  ------- =
}  sqrt(u)


{   
| 2 u^(-1/2) du =
}  

2 u^1/2
-------- = 4 sqrt(u)
 1/2

= 4 sqrt(ln ln x)

Now we have to evaluate that at  x = 3 (no problem) and x = infinity.

But  ln inf = inf, and again ln inf = inf, so this integral DIVERGES, and therefore so does the series.