How can I graphically interpret/picture what it means for a power series to diverge/converge?

Question

A power series is said to converge at a point $x$ if $\lim_{m\to\infty} \sum_{n=0}^{m} a_{n}(x-x_{0})^{n}$ exists for that x and it also has radius of convergence about $x_{0}$. How can I picture what is going on when it says that a power series has a radius of convergence about a point $x_{0}$ and that it can converge to another point, say $x_{1}$, within this radius? And how does that mean that it can't converge for all other values of $x$ outside the interval?

Answer 1

To better understand power series let us look back on just regular infinite series.

When you have an infinite series problem you are given a sequence (an), but instead of analyzing the individual points (x,y) and most likely plotting them you are adding those y values up to get a sum.

Let me give you an example. If you lived forever and I gave you a dollar each day you would have an infinite amount of money as time goes to infinity. However, what if I gave you a dollar on the first day and then half a dollar on the next day and then continued with this process?

If the amount that you keep adding stays the same or increases then the infinite series will diverge and go to infinity, but if the amount that is being added decreases then it will converge.

Now that we understand simple convergence and divergence for infinite series let us now talk about power series.

The only difference between infinite series and power series is that there is a variable x that is put in the sequence that you are summing up.

Because there is a variable that means that different values can be plugged into it. When you plug a value say x=3 into the sequence then it changes the sequence. So in other words you have many many different sequences that you are trying to find the infinite sum of. Some of these sequences will diverge and others will converge.

It turns out that the particular sequences that converge all converge on a certain continuous interval. The center of this interval is called the radius of convergence.

This means that any x value that lies within this interval will make the sequence converge and anything outside of the interval will diverge. This radius is also unique. There cannot be two different intervals of convergence on the x axis for a particular sequence.

For the interval of convergence the interval itself may include its endpoint or not include its endpoints.

Answer 2

A graphical explanation might be difficult to obtain past the $(x-x_0)^3$ term, but the radius of convergence can be understood with reference to the $\textbf{Ratio Test for Convergence}$.

In order for the series to converge for a particular value of $x$ it must be the case that

\begin{equation} \lim_{n\to\infty}\left\vert \dfrac{a_{n+1}(x-x_0)^{n+1}}{a_n(x-x_0)^n}\right\vert<1 \end{equation}

This requirement simplifies to

\begin{equation} \left\vert x-x_0 \right\vert\lim_{n\to\infty}\left\vert \dfrac{a_{n+1}}{a_n}\right\vert<1 \tag{1} \end{equation}

Let ${\displaystyle\lim_{n\to\infty}\left\vert\frac{a_{n+1}}{a_n} \right\vert=c\ge0}$.

If $c=0$ then inequality ($1$) is satisfied for all $x\in\mathbb{R}$.

But if $0<c<\infty$ then inequality ($1$) becomes

\begin{equation} \left\vert x-x_0 \right\vert<\dfrac{1}{c} \end{equation}

so we are guaranteed convergence for $x\in\left(x_0-\frac{1}{c},x_0+\frac{1}{c} \right)$ and denied convergence for $x\notin\left[x_0-\frac{1}{c},x_0+\frac{1}{c} \right]$. ($^*$)

($^*$) Depending upon the particular series one may also obtain conditional convergence at one or both of the endpoints of the interval of convergence.