Meaning of convergence of series in a radius

Question

I know this might sound quite an odd question but I'll try to explain myself. Previously in calculus I understood the meaning of a series to be "converging" like $\sum_{n=1}^\infty \frac{1}{n^2}$ because I understood what it meant, which is that if you keep on summing terms to terms like so $$\frac11+\frac 14+\frac 19+\frac{1}{16}+\dots$$ it would eventually converge to a number that is well known. But I've been learning now complex analysis and never really stopped to understand what it actually means that a series has a radius of convergence like $\sum_{n=0}^\infty z^n=\frac{1}{1-z}$ has the radius of convergence of 1, and that for $|z|<1$ it converges but for $|z|>1$ it diverges, what does it mean?

And yes, I have read several questions here but none satisfied myself fully, also other sources like my book don't really explain this well, they just focus on how to find that radius of convergence, not it's meaning

Answer 1

The difference is that the series $$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\ldots$$ involves adding numbers, while $$\sum_{n=1}^\infty z^n=z+z^2+z^3+\ldots$$ does not. Because $z$ is not a number. It is a variable. So for the first series I mentioned above, you can ask "does it converge"? and that question makes sense with a single yes/no answer. For the series $\sum_{n=1}^\infty z^n$, you can ask "does it converge"? But that question doesn't make sense with one yes/no answer, because you could get different yes/no answers for different choices of $z$. For example, if we take $\sum_{n=1}^\infty z^n$, plugging in $z=1/2$ yields $$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots=1,$$ exactly as you understand convergence. But if we plug in $z=2$, we get $$2+4+8+\ldots,$$ which diverges. So for something like $\sum_{n=0}^\infty c_n(z-a)^n$, instead of asking "does it converge?", we ask "for which $z$ does it converge"? The answer to this question will be a set of $z$s. That set is always a disk centered at the point $a$, and so it makes sense to ask what is the radius of that disk. (The radius can be $0$, in which case the "disk" is really just the point $a$, and the radius can be $\infty$, in which case the "disk" is the whole complex plane).

When we write, for example, $$e^z=\sum_{n=0}^\infty \frac{z^n}{n!},$$ you may be confused about the meaning, because you are thinking of the terms being added until they "eventually converge" to a number (not a function). But all this means is that for each particular $z$ value within the disk of convergence, when I evaluate the series with that specific number, the right side will converge to the function on the left side evaluated at that point, eg $$e^2=\sum_{n=0}^\infty \frac{2^n}{n!}.$$ The right side is a series of numbers and it converges to $$1+\frac{2}{1!}+\frac{4}{2!}+\frac{8}{3!}+\ldots=e^2\approx 7.38905609893.$$

Answer 2

Let't first clear up what "converge" and "diverge" mean.

When a series converge, it means that as the upper bound of the summation approaches infinity, the summation will go towards some value. Rigorously, in terms of limits, if your summation is defined as $S_n=\sum_{k=1}^na_k$ for some sequence $a$, then $\lim_\limits{n\to\infty}S_n=L$ for some finite value $L$.

When a series diverges, it simply means that the limit defined above does not exist.

Now let's discuss about interval of convergence. What an interval of convergence of a series means is essentially: what numbers can be substituted into the summation in order to make the summation converge?

Let's take your geometric series example: $$\sum_{n=0}^\infty z^n=\frac1{1-z}$$

Let's try setting $z$ to be a specific value, like $0.5$. So the series becomes: $$\sum_{n=0}^\infty (0.5)^n$$ We can see that this converges in a variety of ways, but we can do an integral test for example, and see that since $$\int_0^\infty(0.5)^xdx=\frac1{\ln2}$$ i.e. the integral converges, then the corresponding series will also converge due to the integral test.

But now let's take $z=2$, for instance. Now the series becomes: $$\sum_{n=0}^\infty 2^n$$ Because $\lim_\limits{n\to\infty}2^n$ does not exist (it tends towards infinity), we can say that this summation diverges due to the divergence test.

So we can see that for some values, such as $z=0.5$, the series will converge, while for other values, such as $z=2$, the series will diverge.

The interval of convergence is a range of $z$ values such that the series will converge. In this case, it can be found that the series will converge only if $|z|<1$, and will diverge for all other values of $z$. Because of reasons I won't go into here, it turns out that the interval of convergence for any series is either a single continuous interval, empty, or contains all real numbers. Specifically, it won't be a union of disjoint intervals.

As for the radius of convergence, it is essentially just the length of this interval divided by $2$. Because the interval of convergence won't be a union of disjoint intervals, it will always be possible to find some "center" of this interval, and from there, find the radius of the interval of convergence.