Integral Test

[turophile]

I'm working on problems where I need to decide whether a series converges or diverges using the integral test. I've come across a series with a function I can't figure out how to integrate:

? 1/(ln x)[sup:34jfprlw]3[/sup:34jfprlw]

Is there a trick with this one where I should be able to see whether the series converges or diverges without actually evaluating the improper integral? Should I compare it with another (easier to integrate) series? Any help getting started would be appreciated.

 

[galactus]

This is not integrable in the usual sense (elementary methods).

This is known as the Logarithmic Integral. It has its own special name.

\(\displaystyle L_{i}(x)=\int_{2}^{x}\frac{1}{ln(u)}du\)

You see this with regards to higher math and advanced functions.

So, per Mathematica, \(\displaystyle \int\frac{1}{ln^{3}(x)}dx=\frac{1}{2}\left(L_{i}(x)-\frac{x(ln(x)+1)}{ln^{2}(x)}\right)\)

Incidentally, it diverges. \(\displaystyle \int_{2}^{\infty}\frac{1}{ln^{3}(x)}dx=\infty\)


Are you sure it is not

\(\displaystyle \sum_{k=2}^{\infty}\frac{1}{kln(k)^{3}}\)

That is a different matter. It can be integrated. Besides, this is a common problem in most calc texts when one gets to the integral test

 

[turophile]

I double checked, and the series in the problem is ? 1/(ln x)[sup:1adkg483]3[/sup:1adkg483]. Up to this chapter, my book has covered the comparison test, the ratio test, the ratio-comparison test, and L'Hopital's Rule, so I'm going to try some combination of those with the integral test to see where it takes me.

 

[galactus]

You can just use a comparison test.

\(\displaystyle \frac{1}{(ln(k))^{3}}>\frac{1}{k}\)

Since the harmonic series, \(\displaystyle \sum\frac{1}{k}\), diverges so does:

\(\displaystyle \sum_{k=2}^{\infty}\frac{1}{(ln(k))^{3}}\)

\(\displaystyle \frac{1}{(ln(k))^{3}}>\frac{1}{k}\) for large k.

Sorry to ramble in the last post, but I thought you HAD to us the integral test. Not a good choice for this one.

 

[turophile]

Going a bit further into later problems in this chapter, it's become clear to me that the authors wanted to demonstrate when *not* to use the integral test by giving this series to be evaluated for convergence/divergence *before* and *after* introducing Cauchy's condensation test.

If I'm using the condensation test rightly:

The general term of the condensed series is a[sub:3g6t7w19]n[/sub:3g6t7w19] = 2[sup:3g6t7w19]n[/sup:3g6t7w19]/(n * ln 2)[sup:3g6t7w19]3[/sup:3g6t7w19] = 1/(ln 2)[sup:3g6t7w19]3[/sup:3g6t7w19] * 2[sup:3g6t7w19]n[/sup:3g6t7w19]/n[sup:3g6t7w19]3[/sup:3g6t7w19]. Applying L'Hopital's Rule a few times, I'm looking at the limit as n approaches infinity of 1/(ln 2)[sup:3g6t7w19]3[/sup:3g6t7w19] * [n * (n - 1) * (n - 2) * 2[sup:3g6t7w19]n - 3[/sup:3g6t7w19]]/6, which diverges. Thus so does ? 1/(ln x)[sup:3g6t7w19]3[/sup:3g6t7w19].

Thanks for the comparison test example, galactus.