# $\sum {a_n}$ be a convergent series of complex numbers but let $\sum |{a_n}|$ be divergent..

I am stuck on the following problem that says:

Let $\sum {a_n}$ be a convergent series of complex numbers but let $\sum |{a_n}|$ be divergent. Then it follows that
a. $a_n \to 0$ but $|{a_n}|$ does not converge to $0$.

b. the sequence $\{a_n\}$ does not converge to $0$.

c. only finitely many $a_n$’s are $0$.

d. infinitely many $a_n$’s are positive and infinitely many are negative.

e. none of the above.

I am not sure which of the aforementioned options is correct . I think option (a) is the right choice but am unable to prove it. Wiki has been of little help.

Can someone explain? Thanks and regards.

• $a_n = (-1)^n/n$ is a counterexample to (a). – André 3000 May 3 '14 at 7:22
• Actually, NO sequence is such that (a) holds. – Did May 3 '14 at 7:25
• @SpamIAm yes.That is true but can you give a counter example by taking $a_n$ to be a complex number. – learner May 3 '14 at 7:26
• @learner Rational numbers are complex numbers! $\mathbb{Q} \subseteq \mathbb{C}$ – André 3000 May 3 '14 at 7:27
• @SpamIAm yes ,I know that $\Bbb R \subset \Bbb C$ but I was just curious. – learner May 3 '14 at 7:29

Let $a_n=u_n+iv_n$ then the series $$\sum_n a_n$$ is convergent if and only if the two series $$\sum_n u_n\quad\text{and}\quad\sum_n v_n$$ are also convergent and the series $$\sum_n a_n$$ is absolutely convergent if and only if the series $$\sum_n (u_n^2+v_n^2)^{1/2}$$ is convergent.

Now from the above definitions and the fact that if the series

$$\sum_n a_n$$ is convergent then the sequence $(a_n)$ is convergent to $0$ we see that none of the above statements is correct. The following series is a good counterexample:

$$\sum_n \frac{(-1)^n}{n}(1+i)$$

The correct option is (e): The proof that if $\sum a_n$ converges, then $a_n \to 0$ is essentially the same in both the real and complex cases (so (b) is not correct). If $a_n \to 0$, then $|a_n| \to 0$ as well, so it's not (a). Easy counterexamples can be found for (c) and (d) using purely imaginary sequences..

1) Is not true. For example, take $a_n = \dfrac{(-1)^n}{n}$, $n > 0$.

2) Also not true. Take the same sequence.

3) Not true. Take $a_n = \dfrac{(-1)^n}{n}$ if $n$ is odd, and $0$ if $n$ is even. Then, $a_n$ converges and $|a_n|$ does not, but there are not finitely many $a_n$'s that are $0$.

4) True. If finitely many $a_n$'s are negative or positive, we can take them out and the series would be convergent, and since all of the would be positive or negative, the absolute value of the sequence would converge too.

For the fourth one, you could prove it more rigorously, but I let that for you.

• Thanks a lot for your time and effort.. – learner May 3 '14 at 7:32