XIV

Consider the geometric series $1+x+x^2+x^3+\dots +x^n+\dots .$

It is convergent throughout the interval $-1$ to $+1$, excluding the end values $x=\pm 1$.

But it is not uniformly convergent throughout this interval. For if $s_n(x)$ be the sum of $n$ terms, we have proved that the difference between $s_n(x)$ and the limit ${\displaystyle \frac{1}{1-x}}$ is ${\displaystyle \frac{x^{n+1}}{1-x}}$. Now suppose $n$ be any given number of terms, say $20$, and let $k$ be any assigned standard of approximation, say $\mathrm{.001}$. Then, by taking $x$ near enough to $+1$ or near enough to $-1$, we can make the numerical value of ${\displaystyle \frac{x^{21}}{1-x}}$ to be greater than $\mathrm{.001}$. Thus $20$ terms will

not do over the whole interval, though it is more than enough over some parts of it.

The same reasoning can be applied whatever other number we take instead of $20$, and whatever standard of approximation instead of $\mathrm{.001}$. Hence the geometric series $1+x+x^2+x^3+\dots +x^n+\dots$ is non-uniformly convergent over its whole interval of convergence $-1$ to $+1$. But if we take any smaller interval lying at both ends within the interval $-1$ to $+1$, the geometric series is uniformly convergent within it. For example, take the interval $0$ to $+\frac{1}{10}$. Then any value for $n$ which makes ${\displaystyle \frac{x^{n+1}}{1-x}}$ numerically less than $k$ at these limits for $x$ also serves for all values of $x$ between these limits, since it so happens that ${\displaystyle \frac{x^{n+1}}{1-x}}$ diminishes in numerical value as $x$ diminishes in numerical value. For example, take $k=\mathrm{.001}$; then, putting $x=\frac{1}{10}$, we find:–- {3} &\text{for$n=1$,}\quad & \frac{x^{n+1}}{1 - x} &= \frac{(\frac{1}{10})^{2}}{1 - \frac{1}{10}} &&= \tfrac{1}{90} = .0111\dots, \

&\text{for$n=2$,}\quad & \frac{x^{n+1}}{1 - x} &= \frac{(\frac{1}{10})^{3}}{1 - \frac{1}{10}} &&= \tfrac{1}{900} = .00111\dots, \

&\text{for$n=3$,}\quad & \frac{x^{n+1}}{1 - x} &= \frac{(\frac{1}{10})^{4}}{1 - \frac{1}{10}} &&= \tfrac{1}{9000} = .000111\dots.

Thus three terms will do for the whole interval,

though, of course, for some parts of the interval it is more than is necessary. Notice that, because $1+x+x^2+\dots +x^n+\dots$ is convergent (though not uniformly) throughout the interval $-1$ to $+1$, for each value of $x$ in the interval some number of terms $n$ can be found which will satisfy a desired standard of approximation; but, as we take $x$ nearer and nearer to either end value $+1$ or $-1$, larger and larger values of $n$ have to be employed.

It is curious that this important distinction between uniform and non-uniform convergence was not published till 1847 by Stokes–-afterwards,

Sir George Stokes–-and later, independently in 1850 by Seidel, a German

mathematician.

The critical points, where non-uniform convergence comes in, are not necessarily at the limits of the interval throughout which convergence holds. This is a speciality belonging to the geometric series.