Convergence of a Series

[SuperDude]

I'm stuck on a problem dealing with convergence. Current tests I know of are: nth term test, test for geometric, telescoping test(alternating series), integral test, comparison test, and the limit comparison test.

The series in question is:

\(\displaystyle \[\sum_{n=2}^{\infty} \[ \frac {ln(n)} {n sqrt(n+1)} \]\]\)

I'm thinking I need to use the limit comparison test because I can't find a suitable value for the comparison test and the rest of the tests are inconclusive.

This was a question on a quiz we took yesterday, and we got them back today and I can't seem to figure out how to do this

A friend of mine put down that it diverges by the LCT with respect to the power series for p = 3/2 and got it right, but I don't see why this would be so. Evaluating this ratio I come up with

\(\displaystyle \lim_{n \rightarrow \infty}\ \[\frac {\[A_{n}\]} {\[B_{n}\]}\] = \infty\)

Which is inconclusive AFAIK since this test can only imply divergence if both A and B diverge (and B converges for p=3/2).

Any help would be appreciated.

 

[pka]

Take another look at the limit comparison test:
\(\displaystyle \L
\begin{array}{l}
A_n = \left[ {\frac{{\ln (n)}}{{n\sqrt {n + 1} }}} \right]\quad \& \quad B_n = \frac{1}{{n^{4/3} }} \\
\lim _{n \to \infty } \frac{{A_n }}{{B_n }} = 0 \\
\end{array}\).

 

[SuperDude]

Can I ask how you came to chose p=4/3 as your test equation?

The funny thing about this is that our teacher told us when we were taking notes that we wouldn't need to know the rules for when the limit comparison returned 0 or infinity so nobody took notes on it and now the only way to solve a question on this quiz seems to be using one of these rules.

And I find it strange that my friend got it right when he said that it diverges(when it converges) and compared it to a value that was inconclusive.

 

[pka]

Can I ask how you came to chose p=4/3 as your test equation?

Not to be flippant, but it is years of experience.

now the only way to solve a question on this quiz seems to be using one of these rules.

That is never true! There are always multiple ways, especially with series.

Moreover, this is in a class of well known convergent series.
If \(\displaystyle p > 1 \Rightarrow \quad \sum\limits_k {\frac{{\ln (k)}}{{k^p }}} \quad \mbox{converges.}\)

The trick is compare with \(\displaystyle B_n = \frac{1}{{n^{\frac{{1 + p}}{2}} }}\) and get \(\displaystyle \lim _{n \to \infty } \frac{{\ln (n)}}{{n^{\frac{{p - 1}}{2}} }} = 0\)


Here is a postscript: A proof by direct comparison.
\(\displaystyle p > 1 \Rightarrow \quad \frac{{P - 1}}{2} > 0\)

\(\displaystyle \frac{{\ln (n)}}{{n^p }} = \frac{{2\ln \left( {n^{\frac{{p - 1}}{2}} } \right)}}{{(p - 1)n^p }} \le \frac{{2\left( {n^{\frac{{p - 1}}{2}} } \right)}}{{(p - 1)n^p }} = \frac{2}{{(p - 1)n^{\frac{{1 + p}}{2}} }}\)

Clearly \(\displaystyle \sum\limits_n {\frac{2}{{(p - 1)n^{\frac{{1 + p}}{2}} }}}\) converges by the p-test.

 

[SuperDude]

haha, alright, thanks.

Of course, I realize that there's multiple ways of solving these problems, I was speaking out of frustration. Still, I wonder how he expected us to solve this question with the materials he has given us. Even if I had decided to test p=4/3 on test day I wouldn't have known that it meant convergence: it wasn't until after the test that I looked up the other two rules for convergence (for 0 and infinity) because he told us we wouldn't need to know.

I rather doubt he expected us to know that it's a common divergent series and it seems unlikely that anyone in the class would come up with a direct comparison like you did given that we've only had 1 week of experience with the infinite series and all the problems we've been given up to this point have been relatively formulaic in approach. Perhaps I'm missing something still...

And there's still the matter of why my friend got it marked right when he was obviously wrong...

I plan to ask my teacher what he was expecting when we return to school, but since I'm on spring break I'm just going to forget about it for now.

Anywho, thanks for all your help!