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I want to check the convergence of this series $$\sum_{n=2}^{\infty} \frac{(-1)^n}{n\cdot \ln^{7/5}(n)}.$$

When I have this kind of series I pick the positive one and check the tests on it.
$$\sum_{n=2}^{\infty} \frac{1}{n\cdot \ln^{7/5}(n)}$$ I can say that $$n>\ln(n) \Rightarrow \frac{1}{n} < \frac{1}{\ln(n)}$$

My question is if I have chosen the right series for this test.

Ofir Attia
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    If you have an alternating series, then it is naturally to think about alternating series test and check if it works. – Mhenni Benghorbal Jul 26 '13 at 10:48
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    It is natural, but not legitimate — checking, even very quickly, absolute convergence should always be the first idea. If it fails, then* conditional convergence becomes the natural thing to check. – Clement C. Jul 26 '13 at 11:05
  • what would you do if the series was $\frac {-1^k}{k}$? – jimjim Jul 26 '13 at 11:19
  • Observe that it does not converge absolutely, and then use the alternating series test to show it converges conditionally. The first part would take approximately 2 seconds, and I would know for sure — in case I need to work further on my series, eg to show something else — that any result requiring an absolutely convergent series does not apply to my case, so that I'd need to find other ways to deal with it.. – Clement C. Jul 26 '13 at 11:22
  • @Clem Why is checking monotonicity in view of alternatingness not a "legitimate" first move? Why "should" checking a stronger condition than convergence be the first thing to check when verifying mere convergence? Also, I don't think checking absolute convergence here would take someone unacquainted with summands of the form $n^{-\alpha}(\log n)^{-\beta}$ "$2$ seconds" to check; I think it would take longer than the alternating series test. Often we do desire strongest forms of claims, especially with little extra cost, but we're under no moral obligation to desire the strongest forms. – anon Jul 28 '13 at 00:23
  • @julien: I know the OP's series converges absolutely (cf. answer below); I was referring to Arjang's question. (@anon: your argument does make sense, albeit I'd most likely stick with my POV on that...) – Clement C. Jul 28 '13 at 13:31
  • @ClementC. Oops, apologies. To avoid confusion, you can add @ specificuser at the beginning of your comment. – Julien Jul 28 '13 at 13:33

4 Answers4

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Hint: Use alternating series test.

Mhenni Benghorbal
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  • I know its alternating series, I can use Integral test? becuase I see the function and her derivative? – Ofir Attia Jul 26 '13 at 10:14
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    @OfirAttia: Read the statement of the integral test and see if you can apply it to an alternating series. – Mhenni Benghorbal Jul 26 '13 at 10:18
  • after I read it I cant, but I take the positive series so there is no problem right? – Ofir Attia Jul 26 '13 at 10:22
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    @OfirAttia: Use the alternating series test. It is the suitable one for your case. Here is a case where you can use the integral test. – Mhenni Benghorbal Jul 26 '13 at 10:27
  • @OfirAttia: Here is another integral test case. – Mhenni Benghorbal Jul 26 '13 at 10:39
  • The alternating series test will only show conditional convergence — while (this is stronger) the series is absolutely convergent. – Clement C. Jul 26 '13 at 10:48
  • @ClementC.: He was asked to test the convergence of the series and I provided him with a suitable test which is a straightforward one. – Mhenni Benghorbal Jul 26 '13 at 10:58
  • @Downvoter: What's the down vote for? Can you tell us what's wrong with this answer? – Mhenni Benghorbal Jul 26 '13 at 10:59
  • I agree — but in general, when asked about the convergence of a series, the implicit question is to find the strongest form of such convergence (so that other theorems/methods of computations can then be applied either to prove something else or figure the sum). Jumping to the alternating series test because it applies, without first looking at absolute convergence, is usually not a good thing for this reason (one should always start by checking if the simplest and most stringent form of convergence applies; which is clear here because of the theorem aforementioned). – Clement C. Jul 26 '13 at 11:02
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In fact, here your series converges absolutely since we know that $$ \sum_{n=2}^{+\infty} \frac{1}{n^\alpha (\ln n)^{\beta}}<+\infty \Leftrightarrow \left[(\alpha>1)\text{ or }(\alpha=1\text{ and }\beta >1)\right]$$ You can prove this with integral test. And conclude since the absolute convergence implies the convergence.

user37238
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Edited : thanks to Did

There is no need for any test, the terms of series tend to $0$ monotonically alternating sign, that is sufficient that series converges. (There was nothing regrading the absolute convergence in the question, so why bother with it?)

More over the limit of the series L is :

$$\frac{(-1)^2}{2\cdot \ln^{7/5}(2)}+\frac{(-1)^3}{3\cdot \ln^{7/5}(3)}<L<\frac{(-1)^2}{2\cdot \ln^{7/5}(2)} $$

jimjim
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    "the terms of series tend to 0 alternating sign, that is sufficient that series converges"... Actually that is not. – Did Jul 27 '13 at 17:24
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Hint: For $a_n=\frac{1}{n^\alpha \ln^\beta n}$ ($n\geq 2$), the positive series $\sum a_n$ converges if

  • $\alpha > 1$; or
  • $\alpha = 1$ and $\beta > 1$

(and diverges otherwise.)

This'll allow you to see if your series converges absolutely.

Clement C.
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