A Series Breaks Down Far Away
A power series adds up infinitely many powers of x, like 1 + x + x² + x³ + .... How can adding infinitely many terms give a finite value? The secret is distance. Within a radius R of the center the terms shrink fast enough that the sum gathers to one value (it converges). But step outside R and the terms grow instead, so the sum balloons without end and blows up (it diverges). That boundary distance R is called the radius of convergence. The cleanest example is the geometric series 1 + x + x² + ... = 1/(1-x), where R=1. Exactly on the boundary, x=±R, is a delicate spot you must check on its own. In this lesson you will feel, with your own hands, how the partial sum clings inside and leaps outside, how the ratio test pins down R, and how subtle the boundary really is.
Gold is the partial sum SN(x) = 1 + x + ... + xN, gray is the true limit 1/(1-x). Raise the term count N. Inside the interval (-1, 1) the larger N gets, the more snugly gold clings to gray. But on the |x|>1 side it is the opposite: a larger N makes gold leap up and down ever more steeply, fleeing from gray. That is divergence. Move the probe x too. When the dot sits inside the band, S converges to 1/(1-x); outside, it reads as diverges. This single picture holds the whole heart of the lesson: a series lives or dies by distance.
How do you find the radius? The ratio test is the answer. Look at the size ratio of two successive terms, |next term / current term|. For Σ xⁿ this ratio is always just |x|, a constant. The bars are the term sizes |x|ⁿ. Slide x. When |x|<1 the ratio is below 1, so the bars shrink in turn (green, converges); when |x|>1 the bars grow larger and larger (red, diverges). The point where the ratio equals exactly 1, namely |x|=1, is the very border between convergence and divergence, the radius itself. The ratio test stays silent on that border, which the next two blocks handle separately.
Now grab the radius directly on a number line. The series Σ(x/a)ⁿ has radius of convergence R=a. The gold band around the center 0 is the convergence interval (-R, R). Push the radius slider up and the band widens; pull it down and it narrows. Move the probe x. When the point lies inside the band it turns green (|x|<R, converges); when it crosses outside, it turns red (diverges). Convergence comes down to one simple line: is the distance |x| closer than the radius R? Since only the absolute value matters, whether x is positive or negative, the convergence region is always a symmetric interval about the center.
Even with the same radius, fate on the boundary differs case by case. Choose among three series. All have radius 1, yet they behave differently at the boundary x=±1. Σxⁿ diverges at both ends. Σxⁿ/n converges at x=-1, where the terms alternate and partly cancel, but at x=+1 it becomes the harmonic series and diverges. Σxⁿ/n² shrinks its terms fast enough to converge at both ends. The bars trace how the partial sums drift. Settle on one value and it converges; drift on or swing and it diverges. The boundary is where the ratio test lets go, so each series has to be examined on its own like this.
Finally, return to the anchor geometric series. 1 + x + x² + ... = 1/(1-x), but only when |x|<1. Each bar is a term xⁿ, stacking from the top, and the total gathers toward the green dashed line, that is 1/(1-x). Set the ratio x to a positive value like 0.6 and the bars grow steadily shorter on the same side, settling on one value. Make x negative and the bars alternate left and right of the baseline, partly canceling, yet since |x|<1 they still gather to one value in the end. What this picture shows is the essence of a power series: only when the terms shrink fast enough, that is, when the distance lies within the radius, does an infinite sum become finite.