3

I've been having some issues with what to compare it to. I have a hunch it converges. But I just cannot figure out what I can compare it too. Please help. :)

E.C.
  • 89

3 Answers3

10

Hint: A useful heuristic is that for every $\alpha > 0$,

$$\ln n < n^{\alpha}$$

eventually. In particular, one might choose a very small value of $\alpha$, so small that

$$\sum_{n = 1}^{\infty} \frac{n^{\alpha}}{n^{3/2}}$$

converges by studying a $p$-series.

9

Here is another way.

You could also use the fact that, for a decreasing sequence $(a_n)$, the sum $ \sum a_n $ converges if and only if $ \sum (2^n a_{2^n}) $ converges. This is called the Cauchy Condensation Test.

From this, you get rid of the log, since convergence of $ \sum {\log(n) \over n^{3/2}} $ is then equivalent to convergence of $$ \sum (2^n * \log(2^n)* (2^n)^{-3/2}) = \log(2) \sum {n \over 2^{n/2}} $$ and you can probably assume that the latter converges, or at least hopefully you can show it. (If not, then give me a shout!)

Sam OT
  • 6,675
  • 1
    +1 for Cauchy Condensation Test! I taught my class this today. – Squirtle Apr 09 '14 at 22:41
  • 1
    Thanks, and thanks for the edit - I thought that I was forgetting something! :) – Sam OT Apr 10 '14 at 07:17
  • 1
    I don't think this OP is going to up vote or accept an answer being as two scum bags down voted a perfectly valid question. Also (on the level of meta) I become quite irritated when question of the week is some obvious statement or question like, "is it possible to fold an A4 paper in exactly three even pieces" but questions like the one we answered here are voted down. But alas this is our broken world – Squirtle Apr 11 '14 at 05:13
  • I agree completely. I certainly didn't downvote it (in fact, I have upvoted it now to try to restore some balance - perhaps you can do the same?), needless to say since I answered it! – Sam OT Apr 11 '14 at 10:27
  • And yes, I agree that the question of the week should be a less meta maths question, as such; something that has someone has asked because they wanted to know the answer for an actual purpose and someone has given a clear answer. – Sam OT Apr 11 '14 at 10:28
4

Note that for sufficiently large $n$, $\ln(n) \leq n^{1/3}$ and the series

$$ \sum \frac{n^{1/3}}{n^{3/2}} = \sum \frac{1}{n^{7/6}} $$

converges by the $p$-test since $7/6 > 1$. Thus, the series

$$ \sum \frac{\ln(n)}{n^{3/2}} $$

converges by the comparison test.

Ryan
  • 830