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I have to study the character of this series $$\sum_{n=3}^\infty \frac{1}{n(\log(\log n))^{\alpha}}$$

with $\alpha$ a real parameter.

Considering the Cauchy condensation test, the equivalent series is: $$\sum_{n=1}^\infty 2^n\frac{1}{2^n[\log(\log 2^n)]^{\alpha}}=\sum_{n=1}^\infty \frac{1}{[\log(n\log 2)]^{\alpha}}$$

Using the ratio test:

$$\lim_{n\rightarrow \infty} \frac{[\log(n\log 2)]^{\alpha}}{[\log((n+1)\log 2)]^{\alpha}}=\lim_{n\rightarrow \infty} \frac{1}{[\log(\log 2)]^{\alpha}}= \frac{1}{[\log(\log 2)]^{\alpha}}\sim \frac{1}{(-0,36)^{\alpha}}$$

Then when $(-0,36)^{\alpha}>1$ the given series converges, otherwise it diverges

if $\alpha =1$ ,$ -0,36<1$ , diverges

if $\alpha =0$ ,$ 1<1$ , diverges

if $\alpha =-1$ ,$ (-0,36)^{-1}=-2,77<1$ , converges

if $\alpha >1$ ,$ (-0,36)^{\alpha}<1$ , converges

if $\alpha <-1$ ,$ (-0,36)^{\alpha}<1$ , converges

if $|\alpha| <1 , \ne 0$ sometimes it doen't exist $ (-0,36)^{\alpha}$ Can someone help me?

StubbornAtom
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Anne
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3 Answers3

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HINT

Let use limit comparison test with

$$\sum \frac{1}{n}$$

user
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  • should I have to apply the limit comparison test to the given series? – Anne Mar 17 '18 at 10:56
  • @Anne Cuchy condensation is absolutely the right way! to the condensed series we can apply limit comparison test (extended version) with $\sum \frac1n$ – user Mar 17 '18 at 10:57
  • in this way I can say that the series converges for $\alpha \ge 1$ and diverges for $\alpha \le 0$ (and for $0<\alpha < 1 I don't know) $....the suggested solution in my book is convergence for $\alpha > 1$ (1 is excluded!) and divergence otherwise – Anne Mar 17 '18 at 11:16
  • @Anne by LCT we can say that the given series diverges $\forall \alpha$ indeed $$\frac{\frac{1}{[\log(n\log 2)]^{\alpha}}}{\frac1n}=\frac{n}{[\log(n\log 2)]^{\alpha}}\to \infty$$ – user Mar 17 '18 at 11:21
  • but according to th solution, the series converges for $\alpha >1$ – Anne Mar 17 '18 at 11:31
  • @Anne I think there is a typo in the book, if I'm wrong someone else here will give the right answer. How did you apply the ratio test ? How did you obtain the first step? – user Mar 17 '18 at 11:35
  • I applied the ratio test in the text above, Anyway, I think that your solution is right , perhaps there is a mistake in my book – Anne Mar 17 '18 at 11:52
  • @Anne I agree indeed the LCT for the condensed series is quite straightforward and clear, usually when you make also a little mistake you receive tons of downvotes thus it is should be a correct anser :) – user Mar 17 '18 at 11:57
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Since $\sum \dfrac1{n \log(n)}$ diverges and $\dfrac{\log n}{(\log \log n)^a} \to \infty$ for $a > 0$, the sum diverges for all $a$ since $\dfrac1{n(\log \log n)^a} =\dfrac1{n\log n}\dfrac{\log n}{(\log \log n)^a} $.

marty cohen
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You've made a mistake using the ratio test. The ratio of logarithms isn't the logarithm of a difference, it's the other way around (the logarithm of a ratio is the difference of logarithms). Correctly you have: \begin{align} \lim_{n\rightarrow\infty} \frac{\big(\log(n\log 2)\big)^\alpha }{\big(\log((n+1)\log 2)\big)^\alpha} &= \lim_{n\rightarrow\infty} \left(\frac{\log n +\log\log 2}{\log(n+1)+ \log\log 2}\right)^\alpha = \\ &= \lim_{n\rightarrow\infty} \left(\frac{\log n +\log\log 2}{\log n + \log(1+\frac{1}{n})+ \log\log 2}\right)^\alpha = \\ &= \lim_{n\rightarrow\infty} \left(\frac{1 + \frac{\log\log 2}{\log n}}{1 +\frac{\log(1+\frac{1}{n})+ \log\log 2}{\log n}}\right)^\alpha = \\ &= \left(\frac{1 + 0}{1 +0}\right)^\alpha = 1\end{align} which means that the result is inconclusive. You need a different test, for example a comparison test with $\sum\frac{1}{n}$, or another Cauchy condensation test.

Adam Latosiński
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