Summation equivalence

Question

I am having trouble seeing how this summation equivalence holds true:

$\sum_{i=0}^\infty x^i$ = $\frac{1}{1-x}$ if |x| < 1

The only thing I can see where there would be a problem is if x = 1 or x = -1

Im hoping someone can provide me with a clear explanation as I am very confused

Answer 1

Consider the formula for the sum of a finite geometric series: $$ \sum_{i=0}^n x^i = \frac{1-x^n}{1-x}.$$ Now, if we have that the $|x| < 1$, then when we repeatedly multiply $x$ by itself, it gets smaller. This may give some intuition as to why we can then let $n$ tend to infinity in this formula to give $$ \sum_{i=1}^\infty x^i = \frac{1}{1-x},$$ but only when $|x| < 1$, as then $x^n$ tends to 0 as $n$ tends to infinity. We say that 1 is the radius of convergence of the sum - note that $1$ is not included in this radius, so the sum is undefined when $x$ is 1. This should make sense, since $1 + 1 + 1 + \cdots $ intuitively doesn't seem to look like it might end up being finite, right?

Answer 2

I actually thought about this earlier and came up with a somewhat heuristic argument. So we know, for a geometric series with ratio $r$,

$$\sum_{n=0}^k r^n = 1 + r + r^2 + r^3 + ... + r^k = \frac{r^{k+1}-1}{r-1} = \frac{1 - r^{k+1}}{1-r}$$

(Either of the fractions on the right are valid; which you choose is merely a matter of personal taste. You can see their equivalence by multiplying the top and bottom by $-1$. We will be considering the latter however since it's closer to the formula that most people see.)

Consider the limiting behavior of this as $k \to \infty$, i.e. as we accumulate infinitely many terms. What happens? Well, we have...

$$\sum_{n=0}^\infty r^n = \lim_{k \to \infty} \frac{1 - r^{k+1}}{1-r}$$

Well, suppose $r = 1$. Well that's dumb, the limit's undefined, so it's really hard to observe anything there. (Of course, we can show that the limit would be infinity by instead considering the limit of the partial sums.)

Suppose $|r| > 1$. Then clearly, owing to the $r^{k+1}$ term, the fractions increases in magnitude without bound, i.e. diverges. So not much there.

But what if $|r| < 1$? Then $r^{k+1} \to 0$ - it shouldn't be difficult to convince yourself of that much.

Thus, we say, for $|r|<1$,

$$\sum_{n=0}^\infty r^n = \frac{1}{1-r}$$

and for $|r| \geq 1$, it diverges.