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Prove that $\sum \limits_{n=2}^{\infty}{\dfrac{1}{n\ln (n)}}$ diverge

I know it's a well-known series, but all the proofs I've seen are based on the integral test and Cauchy condensation test. I need to prove it using only the following tests: Direct/limit Comparison, Root, D'Alembert, Leibniz, since they are all I have studied so far. Regards and thanks a lot.

Git Gud
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Ab urbe condita
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    It is indeed very easy to do it with those two tests. I'm thinking about doing it with only what you've asked, but it looks hard ; are you sure you're not allowed to use more? Series with logarithms at denominator are often attacked with Cauchy condensation. You could try $\sum_{n} \frac 1{n \log \log n }$ with that test too. – Patrick Da Silva Jun 30 '13 at 22:16
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    If you look at the proof of the Condensation test, you'll see it's really a Comparison test of sorts. Look at the proof here and try to adapt it to your particular series. – David Mitra Jun 30 '13 at 22:17
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    Perhaps one can use $n\ln n\sim p_n?$ – L. F. Jun 30 '13 at 22:17
  • @L. F. If he can't use comparison test, do you really think he can assume that the sum of reciprocals of primes diverge?... lol. – Patrick Da Silva Jun 30 '13 at 22:18
  • Convenient link to other math.SE thread (Cauchy condensation and integral test used) – Zev Chonoles Jun 30 '13 at 22:19
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    @PatrickDaSilva He said explicitly that he can use the comparison test - I don't see your point. It was only a suggestion anyway... – L. F. Jun 30 '13 at 22:20
  • @L.F. Sorry I said things wrong ; I meant if he can't use the integral test. Now I can't edit my comment. sigh. My point is that if he's not allowed to use a hammer to hit that nail, don't give him the most recent Black&Decker super tool to do it. – Patrick Da Silva Jun 30 '13 at 22:24
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    Well: root, d'Alembert y Leibnitz don't work, so there's only the comparison tests (either direct orwith limit)...it doesn't look promising... – DonAntonio Jun 30 '13 at 22:27
  • First, you might try to prove that $\sum_{n=1}^\infty\frac{1}{n^s}$ converges for $s>1$ using your "allowed" tests for inspiration. The difficulties seem similar. – Alex R. Jun 30 '13 at 22:27
  • Thanks for the suggestions, I think I have no choice but to use Condensation test, although I haven´t studied it. @Alex I did prove the generalization of the harmonic series (actually I saw it in a book), but I don´t get your point. – Ab urbe condita Jun 30 '13 at 22:35
  • @DavidMitra You are right, thanks, I have just read the proof of the condensation test, and I realized my teacher used something similar to prove the Harmonic series diverge... – Ab urbe condita Jun 30 '13 at 23:14
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    Though I think it best to just simply show the partial sums are unbounded, as mm-aops does in his answer, if you must: Let $a_n=1/(n\ln n)$. Set $b_3=b_4=a_4$, $b_5=b_6=b_7=b_8=a_8$, $\ldots,$. Use the normal Comparison Test, comparing with $\sum b_i$. – David Mitra Jun 30 '13 at 23:14

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group the terms into blocks of size $2^n$. sum in each block will be bigger than $$ 2^{n} \frac{1}{2^{n+1} \log{2^{n+1}}}$$ so it will be something like the series $\frac{1}{n(2ln(2)} $ which diverges (hope you can use this fact)

mm-aops
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  • Please check your posts before you "send" them...and you seem to be using the condensation test, which isn't allowed. – DonAntonio Jun 30 '13 at 22:31
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    it's kinda like the proof of condesation test goes, but I don't use it explicitly and I suppose that's what the problem is about, he's not allowed to use it because he hasn't studied it yet, so I gave a direct proof for the series in question. – mm-aops Jun 30 '13 at 22:36
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    This is probably the simplist thing to do. It's exactly Oreseme's proof that the Harmonic series diverges; except that the "blocks" of the OP's series are compared to the terms of the Harmonic series. (Of course, this is the Condensation Test. But as I mentioned in my comment above, the Condensation test is just a somewhat fancy Comparison test.) – David Mitra Jun 30 '13 at 22:36
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    And, incidentally, it's not hard to write down an explicit series here and use the (normal) Comparison test (in every block, replace every element with the last term in the block). I see no reason to do this though. – David Mitra Jun 30 '13 at 22:42
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    Seems like direct confrontation of size. Maybe a little better to show directly in this way that $\sum \frac{1}{n\log_2 n}$ diverges. – André Nicolas Jun 30 '13 at 23:12