2

I used Gauss's test and found that we need $(p-1)\ln(n)+q>0$ for this sum to be convergent, which is correct for $p>1$. For $p=1$, I get that $q>0$ is sufficient, which I know is not true, since $\sum \frac{1}{n\ln(n)}$ diverges. Using the integral test, I see that I need $q>1$ when $p=1$ for this sum to be convergent. What's wrong with Gauss's test? I think the problem could be due to the approximation I used with Gauss's test.

Here is an image showing my work

And here is the work, typeset:

$$\frac{a_n}{a_{n+1}} = \frac{(n+1)^p(\ln{(n+1)})^q}{n^p(\ln{n})^q} = \frac{n^p(1+\frac{1}{n})^p}{n^p}\frac{(\ln{n} + \ln{(1+\frac{1}{n})})^q}{(\ln{n})^q}$$ $$= \left(1+\frac{1}{n}\right)^p\frac{(\ln{n} + \frac{1}{n})^q}{(\ln{n})^q} = \left(1+\frac{p}{n}\right)\frac{(\ln{n})^q(1 + \frac{1}{n\ln{n}})^q}{(\ln{n})^q}= \left(1+\frac{p}{n}\right)\left(1 + \frac{1}{n\ln{n}}\right)^q$$ $$= 1 + \frac{p + \frac{q}{\ln{n}}}{n} + \cdots$$ $$p + \frac{q}{\ln{n}} > 1 \Leftrightarrow (p-1)\ln{n} + q > 0$$

Was my approximation invalid? Or does Gauss's test not apply to this series for some reason?

jackson
  • 1,017
user219075
  • 21
  • 3
    Better try the Cauchy condensation test. – Hyperbolic PDE friend Apr 15 '21 at 08:54
  • Also: https://math.stackexchange.com/q/2693749/42969 – Martin R Apr 15 '21 at 09:13
  • 1
    Thanks. I see how to solve the convergence, I'm still wondering if Gauss's test just doesn't work well with this sum. – user219075 Apr 15 '21 at 09:34
  • 1
    I don't think this question should have been closed, as it is not a duplicate. It looks to me like OP is aware of the convergence properties of the series, and came to the correct conclusion using the integral test (which is easier to apply than Cauchy condensation). However, there is a question about Gauss's test, and whether it applies to this problem or has been used incorrectly. That question is not answered in any linked post, and no one has offered this new user any advice on how to clarify the scope of their question in the body, to avoid looking like a duplicate. – jackson Apr 15 '21 at 10:52
  • I'm just pointing out no one tried making that suggestion before closing the question, although this is OP's first post – jackson Apr 15 '21 at 11:33
  • @Jackson It's been reopened. I hope that provides some redemption. Thank you for the comments. – Sarvesh Ravichandran Iyer Apr 15 '21 at 17:34
  • @user219075: What's the statement of Gauss's test that you are using? I know of two basic versions of the test, both given in this question: https://math.stackexchange.com/questions/746871/gauss-test-for-convergence, and which version you are using has a strong influence on the answer to your question – jackson Apr 15 '21 at 17:46
  • @Jackson Thanks, I used the second one in your link, where I ignore all higher order terms of 1/n and looking for A>1. – user219075 Apr 18 '21 at 04:09

1 Answers1

1

This is a Bertrand's series, and it is known to be convergent if & only if $p>1$ or $p=1$ and $q>1$, the latter case being proved very simply by the integral test, the other cases by comparison with a Riemann $p$-series.

Bernard
  • 179,256