Are These Infinite Series Convergent or Divergent?

[horsecandy911]

Homework Statement

I have been straining to find convergence or divergence of a few infinite series. I have tried everything I can think of; ratio test, root test, trying to find a good series for comparison, etc. Here are the formulas for the terms:
#1
1
-------------
(ln(n))^ln(n)

#2
nth root(2) - 1

#3
(k*ln(k))
---------
(k+1)^3

Homework Equations

The Attempt at a Solution

For #1, I noticed that it is equivalent to:
(1/ln(n))^ln(n)
That is, we can raise the 1 on top to the ln(n) power also. Since 1/ln(n) < 1, this sort of resembles a convergent geometric series, but I am unsure how to prove convergence

For #2, the limit of terms approaches 0, so nth-term divergence test does not help us. I tried the ratio test but couldn't evaluate the limit of the resulting ratio; same for the root test. Tried Limit Comparison Test with nth root(2) for my second series, but got that the limit of the ratio was 0, which is inconclusive. I can't tell whether it diverges or converges.

For #3, I think that lnk<k^.5 for large k and (k+1)^3 is greater than k^3, so the terms are less than those of k^1.5/k^3 which = 1/k^1.5, which is a convergent p-series. So by the comparison test this would converge, but I am not sure whether I can use that ln(k)<k^.5 or even if its true.

Thanks for reading and helping!

 

[MathematicalPhysicist]

For the first one try this:
http://en.wikipedia.org/wiki/Cauchy_condensation_test
So the question is does 2^n 1/( ln 2 n^(n ln 2)) converge?
(try to bound 2/n^(ln 2) by some constant which is smaller than 1).

For the second:

2^(1/n)-1 > = 1/n
show this by taking log: 1/n >= log (1+1/n), which is like showing that x>= log(1+x).

The third, I let someone else chip in.

 

[horsecandy911]

Thanks for the help!

I think that my solution to #3 is adequate. I'm not sure how to prove that k^.5 > ln(k), but I can see it from the graphs. I also found this alternative for #1:
http://answers.yahoo.com/question/index?qid=20090318051540AA00Lom

So that is the solution to all the problems.

 

[grey_earl]

If you want to prove \sqrt k > ln k, that you can do by taking derivatives. For k = 1, 2, 3, 4 it is obviously true, and the derivative of \sqrt k is 1/(2 \sqrt k) > 1/k = (ln k)' for k >= 4. So the square root grows faster than the logarithm, and hence must always be greater than it.