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I wish to show $\sum_{n=2}^{\infty}\frac{1}{n\ln n}$ diverges. I initially wanted to use the comparison test, but couldn't come up with a series that is obviously less than $\frac{1}{n\ln n}$ that diverges.

So I moved on to the integral test. The problem here is that I need to show that $f(x)=\frac{1}{x\ln x}$ is continuous on $[2,\infty)$, using $\epsilon$-$\delta$ definition of continuity. I've been given theorems that allow me to just assert that $\ln x$ is continuous on the interval, as I know $1$ is continuous, $x$ is continuous, and so if $\ln x$ is continuous then as we have a composition of continuous functions, $\frac{1}{x\ln x}$ will be continuous on $[2,\infty)$. The trouble I'm having is showing $\ln x$ is continuous.

I again got stuck doing this. And now it feels like I've completely over-complicated things. I need to be rigorous when showing that this series diverges, but we've only been given a certain amount of tools to use. We can't use Cauchy's condensation test, and if I wish to use the integral test I have to show that $f(x)$ is monotone (easy) and also that $f(x)$ is continuous (and the only tool we have for that is $\epsilon$-$\delta$).

I've seen that there are very similar questions to this on the site, but they don't particularly help in my case.

Have I missed something? Thanks for your time.

EDIT: Thank you everyone for your help. Much appreciated!

nooooooooo
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    Integral test is the way to go IMO. What's the derivative of $g(x)=\ln(\ln x)$? Continuity and monotonicity are non-issues, because both $x$ and $\ln x$ are monotonous and positive in your range. No need to go to $\epsilon-\delta$ nitty-gritty. – Jyrki Lahtonen Apr 17 '16 at 05:38
  • THe only tool you have to prove continuity is by the $;\epsilon-\delta;$ definition? And you're already doing infinite series? This doesn't fit into my standard order of things in basic real analysis... – DonAntonio Apr 17 '16 at 05:41
  • Yeah, I'm a little confused. What should have we been introduced to prior to infinite series? – nooooooooo Apr 17 '16 at 05:45
  • How have you defined $\ln x$, and what properties of it do you know? For example, if you know it's differentiable you get continuity for free... –  Apr 17 '16 at 05:46
  • We haven't begun differentiation yet, so am not sure whether I'd be allowed to whip that out. @JyrkiLahtonen Oh! So it should be fine without $\epsilon$-$\delta$? I'm just a bit nervous. Our last piece of assessment was brutally marked if we weren't absolutely rigorous. Ah well! What's a few marks, eh? I'll try reason it out as hole-proof as I'm able I guess! – nooooooooo Apr 17 '16 at 05:49
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    @nooooooooo, if you know that $h(x)=1/x$ is continuous for x>0, $f(x)=ln(x)$ is continuous, $g(x)=x$ is continuous, and products of continuous functions are continuous, as well as knowing compositions of continuous functions are continuous (as long as the relevant requirements are met), then the OP knows that $f(x)g(x)$ is continuous, and so $h(f(x)∗g(x))$ is also continuous. Now use the integral test (which will be very quick). (Copied my comment from below, just in case you didn't see it) – Nicholas Stull Apr 17 '16 at 05:54
  • Would love to, but showing $\ln x$ continuous is the issue I ran into while doing this problem. – nooooooooo Apr 17 '16 at 05:55
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    This completely makes no sense in my mind. Perhaps is some new method in teaching somewhere, but how can you be doing series *with the integral test (and thus I presume you already studied integrals!) without first studying differentiation?? – DonAntonio Apr 17 '16 at 05:57
  • I'm sorry if the question is vexing. I appreciate your time nonetheless. I'll have to just reason it out as well as I can. I posted in case I was missing something very simple. – nooooooooo Apr 17 '16 at 05:59
  • Relevant: http://math.stackexchange.com/questions/916657/epsilon-delta-proof-that-ln-x-is-continuous-everywhere-on-its-domain – Nicholas Stull Apr 17 '16 at 06:02
  • @NicholasStull Thank you. – nooooooooo Apr 17 '16 at 06:06
  • Related: Divergence of $ \sum_{n = 2}^{\infty} \frac{1}{n \ln n}$ through the comparison test? – Martin Sleziak Apr 17 '16 at 07:40

2 Answers2

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To show the continuity, perhaps use this method to show continuity, rather than the epsilon-delta approach.

For an arbitrary point $y$ in your set, consider any sequence $x_n$ converging to $y$.

But we see that $f(x_n) \to f(y)$, which, by definition, shows that $f$ is continuous, as required.

User001
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  • This is explicitly excluded in the question. (But it is certainly one of the fastest ways to show this series diverges) – Nicholas Stull Apr 17 '16 at 05:39
  • Ah, I thought the OP meant he tried the test and doesn't think it was applicable. I misread the context. I will delete my answer. Thanks @NicholasStull. – User001 Apr 17 '16 at 05:40
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    Or try modifying it so that it uses the integral test (which is another really quick way for this one)? – Nicholas Stull Apr 17 '16 at 05:41
  • @NicholasStull The integral test is also immediate almost, yet the asker says he can use it only if he first proves continuity of the function by the $;\epsilon-\delta;$ definition... – DonAntonio Apr 17 '16 at 05:42
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    @Joanpemo which I agree is tedious. But if OP knows that $h(x)=1/x$ is continuous for $x>0$, $f(x)=\ln(x)$ is continuous, $g(x)=x$ is continuous, and products of continuous functions are continuous, then the OP knows that $f(x)g(x)$ is continuous, and so $h(f(x)*g(x))$ is also continuous. Now use the integral test (which as you said, is almost immediate) – Nicholas Stull Apr 17 '16 at 05:48
  • @NicholasStull He said only $;\epsilon-\delta;$ definition can be used. Read it by yoursefl. It hink this is absurd, but that's what he wrote. Not "product of cont. functions is continuous", or "quotient when not zero in the denominator..." and etc. – DonAntonio Apr 17 '16 at 05:55
  • @Joanpemo, OP seems to imply that continuity of compositions of continuous functions and products of continuous functions may be used. If the only obstacle is showing that $\ln(x)$ is continuous, perhaps write up the $\varepsilon$-$\delta$ proof that $\ln(x)$ is continuous (which is indeed bordering on the absurd) – Nicholas Stull Apr 17 '16 at 06:00
  • @NicholasStull I did try to write up that proof, but got stuck. – nooooooooo Apr 17 '16 at 06:01
  • I just caught up with reading your comments; I used a sequential characterization of continuity, not the $\epsilon-\delta$ method... – User001 Apr 17 '16 at 06:01
  • @User001 I will try this way as well, and see where it takes me. Thank you :) – nooooooooo Apr 17 '16 at 06:02
  • Sure thing -- it is as "rigorous" as the $\epsilon - \delta$ method. It's very common to use sequential characterizations. Good luck. – User001 Apr 17 '16 at 06:03
  • @nooooooooo, see my last comment on the question (it links to an $\varepsilon$-$\delta$ proof) – Nicholas Stull Apr 17 '16 at 06:04
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Cauchy Condensation Test: If $(x(n))_{n\in N}$ is monotonic for all sufficiently large $n$, then $\sum_{n\in N}x_n$ converges iff $\;\sum_{n\in N}2^n x(2^n)$ converges.

With $x(n)=1/(n\ln n)$ for $n\geq 2,$ we have $2^n x(2^n)=1/(n\ln 2)$ for $n\geq 2.$ Applying the Cauchy test again to $(y(n))_{n\in N}$, where $y(n)=1/(n\ln 2),$ we have $2^ny(2^n)=1/\ln 2.$ And $\sum_{n\in N}1/\ln 2=\infty.$ Hence $\sum_n x_n$ diverges.

This (repeated) test works for constant $k$ for $x_n=n^{-k},$ for $x_n=^*[n (\ln n)^k)]^{-1},$ for $x_n=^*[(n\ln n )(\ln (\ln n))^k]^{-1},$ etc., where, by $=^*$, I mean "equal for all but finitely many $n$."

DanielWainfleet
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