# Thread: Calculus II - Comparison Test(s)

1. ## Calculus II - Comparison Test(s)

I have two similar questions:

1.) Sigma from n=1 to infinity (1 / (sqrt(n) * (ln n)^4))

2.) Sigma from n=1 to infinity ((ln n)^3 / n^2)

I must use the direct comparison/limit comparison tests for these, but I'm not sure how to go about choosing a bn.

My first intuition for #1 was to choose bn to be 1/n, but 1/n is divergent and greater than (1 / (sqrt(n) * (ln n)^4)).

Any help would be appreciated because I'm very lost.

2. since log x is a slowly varying function, you can more or less ignore it in this case

By that I mean the first series will diverge and the second will converge.
NOW, you can either do the direct comp or limit comp tests.

Since the essential p=.5 in the first series you can use any p in (.5,1] say p=2/3.
Compare that series to

$\sum {1\over n^{2/3}}=\infty$ or better yet $\sum {1\over n}=\infty$

Next show that ${{1 \over \sqrt{n} (\ln n)^4}\over {1\over n}}\to\infty$ and you're done

Showing this limit is infinite means your terms are larger hence your series is larger
and we all know that the harmonic series diverges.

3. Same idea with $\sum {(\ln n)^3 \over n^2}$

This converges, BUT it's slightly bigger than $\sum {1\over n^2}$

SO here, pick any p in (1,2) and you win, say p=3/2...

$\sum {1\over n^{3/2}}<\infty$

Finally take either ratio and show that your terms are much smaller than ${1\over n^{3/2}}$

It doesn't matter which ratio you use.
One will go to zero, the other to infinity.
The point is, you found a larger sum that's finite.
Since all the terms in both series are positive your series will converge too.
A direct comp test works just as well.
The only problem is that we may need to ignore the first few terms.