$$\sum^{\infty}_{n=1}\frac{(-1)^n}{n^a\ln n}$$ $$a>0$$

Does the series converge/converge absolutely/diverge ?

I tried to divide to cases and factor the series:

$\sum^{\infty}_{n=1}\frac{(-1)^n}{n^a\ln n}=\sum\frac{(-1)^n}{n^a}\sum\frac{1}{\ln n}$

for $a \le 1$ the series $\sum\frac{(-1)^n}{n^a}$ converges (from Leibniz), and doesn't converge absolutely but $\sum\frac{1}{\ln n} $ diverges. So I guess factoring the series into two is not how to solve this.

Now trying the condensation test:

$\sum^{\infty}_{n=1}\frac{2^n(-1)^{2^n}}{{2^n}^a\ln 2^n} \Rightarrow \frac{2^{n(1-a)}}{\ln 2^n}$

If $a \ge 1$ it goes to infinity.

If $a \le 1$ it also goes to infinity.

I think I'm doing somthing wrong.

Any advice ?

  • $\begingroup$ This is an alternating series. Try out the Leibniz test-en.wikipedia.org/wiki/Alternating_series_test $\endgroup$ – GTX OC Dec 7 '13 at 10:26
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    $\begingroup$ The condensation test is for monotonically non-increasing terms, not for alternating terms. $\endgroup$ – Daniel Fischer Dec 7 '13 at 10:26
  • $\begingroup$ Right, so it does converge for all a. and converge absolutely for $a > 1$ and doesn't converge absolutely for $a \le1$. $\endgroup$ – GinKin Dec 7 '13 at 10:39
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    $\begingroup$ The step where you factor the series is not valid. There is no rule that says that $$\sum_na_nb_n=\sum_na_n\sum_nb_n$$ You would only use the condensation test for the absolute convergence. $\endgroup$ – robjohn Dec 7 '13 at 12:09

The alternating series test says that $$ \sum_{n=2}^\infty\frac{(-1)^n}{n^a\log(n)} $$ converges conditionally since $\frac1{n^a\log(n)}$ monotonically converges to $0$.

For absolute convergence, by comparison to $$ \sum_{n=3}^\infty\frac1{n^a} $$ the series converges absolutely for $a\gt1$.

For $a=1$, the integral test shows that the series diverges absolutely since $$ \int_2^M\frac1{x\log(x)}\,\mathrm{d}x=\log(\log(M))-\log(\log(2)) $$ diverges as $M\to\infty$.

If you can't use the integral test, you can use the condensation test for $a=1$: $$ \sum_{n=1}^\infty2^n\frac1{2^n\log(2^n)}=\frac1{\log(2)}\sum_{n=1}^\infty\frac1n $$ diverges since the harmonic series diverges.

Since the series diverges for $a=1$, the comparison test with $$ \sum_{n=2}^\infty\frac1{n\log(n)} $$ shows that the series diverges absolutely for $a<1$.

  • $\begingroup$ so it converge absolutely for any $a$? $\endgroup$ – guynaa Dec 7 '13 at 11:56
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    $\begingroup$ @guynaa: no. The alternating series test only works for conditional convergence. I noticed the part of the question regarding absolute convergence and added a section on absolute convergence. $\endgroup$ – robjohn Dec 7 '13 at 12:02
  • $\begingroup$ We can't use integrals yet. Can we use that comparison test for $ a \le 1$ ? $\endgroup$ – GinKin Dec 7 '13 at 12:10
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    $\begingroup$ You can use comparison for $a\lt1$, but for $a=1$, you can use the condensation test. $\endgroup$ – robjohn Dec 7 '13 at 12:13

Firstly note that your series should be $$\sum_{n=2}^{\infty}\frac{(-1)^n}{n^a \ln n}$$ since the logarithm function becomes zero at $n=1$. Let us consider the absolute valued series first,

$$\sum_{n=2}^{\infty}\frac{1}{n^a \ln n}$$

Note that,

$$\frac{1}{n^a \ln n}<\frac{1}{n^a}\mbox{ for }n\geq 3$$

Hence by the direct comparison test we have $\sum_{n=2}^{\infty}\frac{1}{n^a \ln n}$ converges when $a>1$. That is $\sum_{n=2}^{\infty}\frac{(-1)^n}{n^a \ln n}$ absolutely converges when $a>1$.

Ir remains to show the case $a\leq 1$. I still haven't come with a method for that. :)

  • $\begingroup$ Can't you just compare it to 1/n for $a \le 1$? $\endgroup$ – GinKin Dec 7 '13 at 11:09
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    $\begingroup$ Well the comparison test only works when all the terms of the series are positive. But in this case you have a alternating series. If you take the absolute valued seres, $\sum_{n=2}^{\infty}\frac{1}{n^a \ln n}$ then using the limit comparison test with $1/n$ is useless because $\sum\frac{1}{n}$ diverges and then we cannot imply anything about our original series. $\endgroup$ – user112535 Dec 7 '13 at 11:21
  • $\begingroup$ Rather than using the above method, I now realize that this could be solved using the alternating series test. :) $\endgroup$ – user112535 Dec 7 '13 at 11:51

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