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I know I have seen something similar and there is a telescoping trick to the convergence but it is eluding me.

Thomas Andrews
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victoria
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  • Use Cauchy condensation theorem. – Nosrati Apr 08 '17 at 02:22
  • Hi Victoria! Using either the Integral Test or "Cauchy condensation", like MyGlasses suggests, you should be able to find an answer. (It does not involve telescoping, though.) – Chris Apr 08 '17 at 02:23
  • Question comes from a (weak) Calculus 2 student. Need to use a teachnique available after 1,5 semensters of calculus. – victoria Apr 08 '17 at 02:23
  • Integral test might work but a bit high level for this guy. – victoria Apr 08 '17 at 02:24
  • This problem happens to be kind of a classic one for using the integral test, as far as I know. Did your instructor not teach it yet? – Chris Apr 08 '17 at 02:25
  • Not my instructor, tutoring student's instructor and no idea since he is very confused. OK integral test it is, thanks. – victoria Apr 08 '17 at 02:28

3 Answers3

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We circumvent using the integral test or its companion, the Cauchy condensation test. Rather, we use creative telescoping to show that the series $\sum_{n=3}^\infty \frac{1}{n\log(n)}$ diverges. To that end, we now proceed.

We will use the well-known inequalities for the logarithm (SEE THIS ANSWER)

$$\frac{x-1}{x} \le \log(x)\le x-1 \tag1$$

Using the right-hand side inequality in $(1)$, we see that

$$\log\left(\frac{n+1}{n}\right)\le \frac1n \tag 2$$

and

$$\log\left(\frac{\log(n+1)}{\log(n)}\right)\le \frac{\log(n+1)}{\log(n)}-1 \tag3$$

Applying $(2)$ and $(3)$ yields

$$\begin{align} \sum_{n=3}^N \frac{1}{n\log(n)} &\ge \sum_{n=3}^N \frac{\log\left(\frac{n+1}{n}\right)}{\log(n)}\\\\ &=\sum_{n=3}^N \left(\frac{\log(n+1)}{\log(n)} -1\right)\\\\ &\ge \sum_{n=3}^N \log\left(\frac{\log(n+1)}{\log(n)}\right)\\\\ &=\sum_{n=3}^N \left(\log(\log(n+1)) -\log(\log(n)) \right)\\\\ &=\log(\log(N+1))-\log(\log(3)) \end{align}$$

Inasmuch as $\lim_{N\to \infty}\log(\log(N+1))=\infty$, the series of interest diverges by comparison.

And we are done!

TOOLS USED: The right-hand side inequality in $(1)$ and summing a telescoping series.

Community
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Mark Viola
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  • Victoria, please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark – Mark Viola Apr 09 '17 at 19:09
  • @chris Neither the integral test nor its companion Cauchy's condensation test is required. We need only apply an elementary inequality and a telescoping series. So, even pre-calculus students are able to handle this one. -Mark – Mark Viola Apr 09 '17 at 19:14
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For a decreasing function $a(x) \ge 0$, there is a theorem known as the Integral Test, which says that $\int_{m}^\infty a(x)$ converges if, and only if, $\sum_{n=m}^\infty a(n)$ converges. This is the (probably best) way of determining convergence or divergence here - you'll want to take $a(x) = \frac{1}{x\ln(x)}$.

Chris
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    Maybe $\ln\ln x$ instead? – Clement C. Apr 08 '17 at 02:36
  • $a(x) = (x\ln(x))^{-1}$? – Neal Apr 08 '17 at 02:41
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    You need more than $a(x)\geq 0$ - you need $a(x)$ is decreasing for $x\geq m$. – Thomas Andrews Apr 08 '17 at 02:46
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    I got it with the integral test; integral of the series is indeed ln(ln n)

    Was hoping one of the cute tricks would work. Student apparently has not yet read the chapter.

     – victoria Apr 08 '17 at 03:05
  • @victoria Well then. No cure for not having read the chapter. – Neal Apr 09 '17 at 12:15
  • Yup. I get them all the time. For this prize, I sent ten sample convergence test examples, and he got mad at me because I hadn't "worked out the steps" to suit his fancy. Then he sent a sample from class which was a root test. Uhh -- OK but there are nine other tests and I just showed you how to use them. Argh. – victoria Apr 13 '17 at 04:29
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The easiest way to do this is use Cauchy...$f(n)$ coverges or diverges so as $2^nf(2^n)$ converges or diverges.But then the latter is $\Sigma$ $1/nln(2)$ which diverges.

CoffeeCCD
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