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(*Lecture 16 of Mathematical Methods II.*)

A series is a sum of, typically, infinitely many terms. It therefore involves two concepts: that of a *sequence* of terms, that brings together the terms to be summed. And the limit of their sums.

Let us start with defining the sequence $(z_n)$. This is the notation for the collection of terms $z_0$, $z_1$, $\ldots$, $z_k$ with $k\in\mathbf{N}$. Such a sequence is called *convergent* if it has a limit. In symbols:

$$(\exists c\in\mathbf{C})(\forall\epsilon>0)(\exists N\in\mathbf{N})(n>N)\Rightarrow(|z_n-c|<\epsilon)$$

A sequence~$(z_n)$ of complex numbers $z_n=x_n+iy_n$ converges to $c=a+ib$ iff the sequence of real and imaginary parts of~$z_n$ converge to the real and imaginary part of~$c$.

From there we are ready to introduce the notion of a series of complex numbers: given a sequence~$(z_n)$, we define another sequence $(s_m)$ such that:

$$s_m=\sum_{n=0}^mz_n\,.$$

Each term is called a partial sum and the sequence $(s_m)$ itself is what is referred to as a *series*.

A convergent series is one who sequence of partial sums converge.

*Theorem:* If $\sum z_m$ converges, then $\lim_{m\rightarrow\infty}z_m\rightarrow0$.

*Proof:* Assuming that $\lim_{k\rightarrow\infty}\sum_k z_k$ converges, and calling $s$ its limit, then since $z_m=\sum_{k=1}^mz_k-\sum_{k=1}^{m-1}$ for all $m$, taking the limit of both sides, and since we assumed the existence of the limit of $\sum_k z_k$, the limit of the difference is the difference of the limits and we find:

$$\lim_{m\rightarrow\infty}z_m=\lim_{m\rightarrow\infty}\left(\sum_{k=1}^mz_k-\sum_{k=1}^{m-1}z_k\right)=s-s=0\,.$$

Therefore, if $z_m$ does not go to zero, it is immediate that the series made out of the corresponding sequence does not converge. Of course while $z_m\rightarrow0$ is a necessary condition for convergence of the series, it is not sufficient. For instance, the harmonic series $\sum_{k=1}^\infty\frac{1}{k}$ diverges.

A series is called *absolutely convergent* if the series of absolute terms converge.

A series which is such that $\sum_k z_k$ converges but $\sum_k|z_k|$ diverges is called *conditionally convergent*. For instance, the series $\sum_k\frac{(-1)^k}{k}$ converges but the sum of absolute terms diverges, as just mentioned.

It is not always easy to prove that a series converges. One chief difficulty is that the limit might not be done in the first place. In this case, one can use the powerful *Cauchy's Convergence Principle* which asserts that a series $\sum z_m$ is convergent iff for every~$\epsilon>0$, there exists~$N$ such
that~$|z_{n+1}+z_{n+2}+\cdots+z_{n+p}|<\epsilon$ for every~$n>N$ and~$p\in\mathbf{N}$.

Another useful property is that provided by the *comparison test*: If $\sum b_n$ is a convergent series such that $|z_n|\le b_n$ for all~$n$, then $\sum z_n$ converges absolutely.

This can be proved thanks to Cauchy's principle. Given that $b_k$ converges, then by this principle, for any $\epsilon>0$, there exists $N$ such that $b_{n+1}+\cdots+b_{n+p}<\epsilon$ for every $n\ge N$ and for all $p\in\mathbf{N}$. From this and from the bounding of $|z_k|$ by the $b_k$, we have transported the Cauchy property to the $|z_k|$, i.e., $|z_{n+1}|+\cdots+|z_{n+p}|<\epsilon$, proving the convergence with the other implication of the Cauchy equivalence.

The geometric series is a good comparison series:

\begin{equation} \tag{1} \sum_{m=0}^\infty q^m=\frac1{1-q} \end{equation}

if~$|q|<1$ and diverges otherwise.

*Ratio Test*: If a series $(z_n)$ has the property that for every~$n$ greater than~$N$:

\begin{equation} \tag{2} \left|\frac{z_{n+1}}{z_n}\right|\le q<1 \end{equation}

the series converges absolutely; if for every~$n>N$,

\begin{equation} \tag{3} \left|\frac{z_{n+1}}{z_n}\right|\ge1 \end{equation}

then the series diverges. These will provide useful criteria for the power series that we will study later.

*Proof:* If (3) holds, then for all $n\ge N$, $|z_{n+1}|\ge|z_n|$ and therefore $|z_{n+1}|>|z_n|/2$ and therefore does not tend to zero, therefore impeding convergence of the series. On the other hand, if (2) holds, then $|z_{N+2}|\le|z_{N+1}|q$, $|z_{N+3}|\le|z_{N+1}|q^2$, that is, by recurrence:

\begin{equation} |z_{N+p}|\le|z_{N+1}|q^{p-1}\,, \end{equation}

from which we deduce:

\begin{equation} |z_{N+1}|+|z_{N+2}|+\cdots\le|z_{N+1}|(1+q+q^2+\cdots)\,, \end{equation}

which proves the absolute convergence of $(z_n)$ from the comparison test (the sequence $(z_{N+1} q^n)$ is convergent, to $|z_{N+1}|/(1-q)$).

If the sequences of the ratios converge to a nonzero value (i.e., if zero is not an accumulation point, so that we can apply the ratio test past a large enough integer), it is convenient to estimate whether a series converge from the previous property:

Calling $L$ the limit of the ratios:

$$\lim_{n\rightarrow\infty}\left|\frac{z_{n+1}}{z_n}\right|=L\,.$$

Then:

- If $L<1$, the series converges absolutely.
- If $L>1$, the series diverges.
- If $L=1$, the series may converge or diverge.

This is an immediate consequence of the ratio test applied to $k_n=|z_{n+1}/z_n|$, that can be made $<q$ for a $n$ large enough.