0
$\begingroup$

Is the following series absolutely convergent?

$$\sum_{n\ge1}{\frac{1+(-1)^nni}{n^2}}$$ where $i$ is the complex number such that $i^2=-1$.

I don't know how to start it, I'm not good with complex numbers.

  • 0
    Did you mean to say $(-1)^n$? – 2017-01-08
  • 0
    @SimpleArt, Yes, and sorry, I missed it the first time. – 2017-01-08
  • 0
    So the second term in the numerator is $(-1)^nni$, just to be clear? – 2017-01-08
  • 0
    @SimpleArt Yes it is – 2017-01-08
  • 0
    Is $(-1)^n/n$ convergent?How about $1/n^2$? – 2017-01-08
  • 0
    @kingW3 yes it is, but how it will help me solve the exercise? – 2017-01-08
  • 0
    @Neacsu Mihai You can split the $1/n^2$ since it converges absolutely and then you're left with the sum $\frac{(-1)^n}{n}\cdot i$ but since $i$ is a constant you can pull it out of the sum. – 2017-01-08
  • 0
    @kingW3 Hm, technically, you are pulling the $(-1)^ni/n$ out of the series, leaving you with $1/n^2$, but that doesn't converge absolutely :P – 2017-01-08

2 Answers 2

2

One may notice that

$$\sum_{n\ge1}\frac{1+(-1)^nni}{n^2}=\sum_{n\ge1}\frac1{n^2}+\sum_{n\ge1}\frac{(-1)^ni}n$$

The first converges due to the Cauchy condensation test and the second converges by the alternating test. However, the second one fails to converge absolutely (again, by Cauchy condensation test), thus, your series converges conditionally.

  • 0
    Splitting the infinite series like this needs justification which I don't think is there. – 2017-01-08
  • 0
    So I don't actualy need to do anything to that $i$, but what if i have $i^n$ instead of $(-1)^n$, will that change the outcome or it will still diverge $\sum_{n\ge1}\frac{(-1)^ni}n$? – 2017-01-08
  • 0
    @Alqatrkapa Hm... and why do you think that? If I recall, justification is needed for the combining of series, not the other way around, though I suppose I may be wrong. – 2017-01-08
  • 0
    @NeacsuMihai For more general statements like that, you would use the [Dirichlet test](https://en.wikipedia.org/wiki/Dirichlet%27s_test), which is a more general statement than the alternating test. – 2017-01-08
  • 0
    @SimpleArt Splitting and combining are the same operation in reverse. For instance, I can split the convergent series $\sum_n 0$ into the two divergent series $\sum_n 1/n$ and $\sum_n -1/n$ term-by-term. This is a trivial example but it illustrates the problem. If you could split the series into two *absolutely* convergent series, then you would get somewhere. – 2017-01-08
  • 0
    @Alqatrkapa Yes, and when one gets two divergent series, that is like indeterminate form. It tells nothing one way or the other. But if they both converge, then it tells you something. – 2017-01-08
  • 0
    @SimpleArt Ok you are right, if you split a series into two (normally) convergent series you have a convergent series. Just had to get over my instinctive fears of splitting series :) – 2017-01-08
  • 0
    @SimpleArt Sorry cuz i have a lot of questions, but I rly want to understand.. So, why should i use Dirichlet test? I mean $i$ is complex so $i^2=-1$ so its an alternating series.. – 2017-01-08
  • 0
    @Alqatrkapa Those are good instincts. See the rearrangement theorem for more fears. – 2017-01-08
  • 0
    @NeacsuMihai It is not alternating in the normal sense. We have our $i^n$ behaving as follows:$$i,-1,-i,1,i,-1,\dots$$and the alternating test only handles things of the following:$$1,-1,1,-1,\dots$$? – 2017-01-08
0

We can calculate the absolute value of a term directly. Recalling that for complex $z$, $|z|^2 = z^*z$ (where $z^*$ denotes the complex conjugate of $z$), the absolute value of the term is

$$\left| \frac{1+(-1)^nni}{n^2} \right| = \frac{\sqrt{1 + n^2}}{n^2}$$

$$\sim\frac{1}{n}$$

and so the series of absolute terms diverges in comparison to the harmonic series.

  • 0
    So it is, at best, conditionally convergent. – 2017-01-09
  • 0
    @martycohen Indeed, see the other answer for a proof that it converges. – 2017-01-09