# Series with complex numbers

#### Lisa91

##### New member
Let's take $$\sum_{n=1}^{\infty} (-i)^{n} a_{n}$$, which is convergent , $$a_{n} > 0$$. What can we say about the convergence of this one: $$\sum_{n=1}^{\infty} (-1)^{n} a_{n}?$$ What can I do with it?

#### MarkFL

Staff member
I would try writing the known convergent series as a complex value in rectangular form where we know the two parameters must also be convergent.

#### Lisa91

##### New member
Do you mean something like this $$\sum_{n=2k}^{\infty} (-i)^{2k} a_{2k}$$? Well, we could use $$\frac{1}{(2k)^{2}}$$ instead of $$a_{2k}$$.

#### MarkFL

Staff member
I meant to write:

$\displaystyle \sum_{n=1}^{\infty}(-i)^na_n=\sum_{n=1}^{\infty}(-1)^na_{2n}+i\sum_{n=1}^{\infty}(-1)^na_{2n-1}=a+bi$

Therefore:

$\displaystyle \sum_{n=1}^{\infty}(-1)^na_{2n}+\sum_{n=1}^{\infty}(-1)^na_{2n-1}=a+b$

$\displaystyle \sum_{n=1}^{\infty}(-1)^na_{n}=a+b$

#### Lisa91

##### New member
I am not so sure whether I got the idea. I divide the series into two parts. It's clear when I write its first terms. But then we take $$a+b$$ - the imaginary and real part. On what basis can we add these two parts?

I guess the main idea is to show that the series $$\sum_{n=1}^{\infty}(-1)^na_{n}$$ is a part of it so it must be convergent but I don't really see it.

#### MarkFL

Staff member
If the given series converges to some complex value, i.e.,

$\displaystyle \sum_{n=1}^{\infty}(-i)^na_n=\sum_{n=1}^{\infty}(-1)^na_{2n}+i\sum_{n=1}^{\infty}(-1)^na_{2n-1}=a+bi$

Then we must have:

$\displaystyle \sum_{n=1}^{\infty}(-1)^na_{2n}=a$

$\displaystyle \sum_{n=1}^{\infty}(-1)^na_{2n-1}=b$

and so:

$\displaystyle \sum_{n=1}^{\infty}(-1)^na_{2n}+\sum_{n=1}^{\infty}(-1)^na_{2n-1}=\sum_{n=1}^{\infty}(-1)^na_{n}=a+b$

#### Lisa91

##### New member
We can write the following ones as:

$$\sum_{n=1}^{\infty}(-1)^na_{2n}$$

$$-a_{1}+a_{3}-a_{5}+...$$

$$\sum_{n=1}^{\infty}(-1)^na_{2n-1}$$

$$-a_{2}+a_{4}-a_{6}+...$$

So we get:

$$-a_{1}-a_{2}+a_{3}+a_{4}-a_{5}-a_{6}+...$$

$$\sum_{n=1}^{\infty}(-1)^na_{n}$$ the series differs a liitle bit.

#### chisigma

##### Well-known member
Let's take $$\sum_{n=1}^{\infty} (-i)^{n} a_{n}$$, which is convergent , $$a_{n} > 0$$. What can we say about the convergence of this one: $$\sum_{n=1}^{\infty} (-1)^{n} a_{n}?$$ What can I do with it?
According to the so called 'Abel's Test' if for a complex sequence $a_{n}$ is $\displaystyle \lim_{n \rightarrow \infty} a_{n}=0$ and the series...

$\displaystyle f(z)= \sum_{n=0}^{\infty} a_{n}\ z^{n}$ (1)

... converges when $|z|<1$ and diverges when $|z|>1$ and the coeffcints $a_{n}$ are positive real numbers monotonically tending to 0, then f(z) converges everywhere on the unit circle except in z=1. Are You in such conditions?...

Kind regards

$\chi$ $\sigma$

#### Lisa91

##### New member
According to the so called 'Abel's Test' if for a complex sequence $a_{n}$ is $\displaystyle \lim_{n \rightarrow \infty} a_{n}=0$ and the series...

$\displaystyle f(z)= \sum_{n=0}^{\infty} a_{n}\ z^{n}$ (1)

... converges when $|z|<1$ and diverges when $|z|>1$ and the coeffcints $a_{n}$ are positive real numbers monotonically tending to 0, then f(z) converges everywhere on the unit circle except in z=1. Are You in such conditions?...

Kind regards

$\chi$ $\sigma$
$$|z|= |0^{2}+(-1)^{2}| = |1|$$ so according to the rule we don't know whether it is convergent or not...

#### chisigma

##### Well-known member
$$|z|= |0^{2}+(-1)^{2}| = |1|$$ so according to the rule we don't know whether it is convergent or not...
If You consider the series...

$\displaystyle \sum_{n=0}^{\infty} a_{n}\ z^{n}$ (1)

... You know that it converges for $z=-i$. That implies that, because $(-i)^{n}$ doesn't tend to 0 with n, is $\displaystyle \lim_{n \rightarrow \infty} a_{n}=0$. You know also that for all n is $a_{n}>0$. Two of the requirement of the Abel criterion are satisfied, so that You can [preliminary...] conclude that the series converges on all the unit circle with the exception of z=1 [so that it converges for z=-1...] if for all n 'large enough' is $a_{n+1}<a_{n}$. Now we have to analyse if the last limitation can be in some way overcomed...

Kind regards

$\chi$ $\sigma$

#### MarkFL

Staff member
We can write the following ones as:

$$\sum_{n=1}^{\infty}(-1)^na_{2n}$$

$$-a_{1}+a_{3}-a_{5}+...$$

$$\sum_{n=1}^{\infty}(-1)^na_{2n-1}$$

$$-a_{2}+a_{4}-a_{6}+...$$

So we get:

$$-a_{1}-a_{2}+a_{3}+a_{4}-a_{5}-a_{6}+...$$

$$\sum_{n=1}^{\infty}(-1)^na_{n}$$ the series differs a liitle bit.
You're absolutely right; what I suggested doesn't work. #### Opalg

##### MHB Oldtimer
Staff member
Let's take $$\sum_{n=1}^{\infty} (-i)^{n} a_{n}$$, which is convergent , $$a_{n} > 0$$. What can we say about the convergence of this one: $$\sum_{n=1}^{\infty} (-1)^{n} a_{n}?$$ What can I do with it?
Under those conditions, the series $\sum_{n=1}^{\infty} (-1)^{n} a_{n}$ need not converge. Suppose for example that $a_n = \begin{cases} 1/n^2 & (n \text{ odd}) \\ 1/n & (n \text{ even}) \end{cases}$. Then $$\sum_{n=1}^{\infty} (-i)^{n} a_{n} = \biggl(i\sum_1^\infty \frac{(-1)^n}{(2n-1)^2}\biggr) + \biggl(\sum_1^\infty \frac{(-1)^n}{2n}\biggr),$$ and both of the bracketed series converge. But $$\sum_{n=1}^{\infty} (-1)^{n} a_{n} = -1+\frac12 -\frac1{3^2} + \frac14 -\frac1{5^2} + \frac16 - \ldots\,.$$ In that series, the negative terms form a convergent series, but the positive terms form a divergent series, and so the whole series must diverge.