So far a series has summed to a number. A power series does
something far more powerful: it sums to a function. Pick a centre
a and a sequence of coefficients
c_0, c_1, c_2, \dots, and form
f(x) \ = \ \sum_{n=0}^{\infty} c_n (x - a)^n \ = \ c_0 + c_1(x-a) + c_2(x-a)^2 + \cdots.
For each fixed x this is just a numerical series — it either
converges or it does not. The set of x where it converges is the
domain of the function f. So a power series is an
infinite-degree polynomial, and whether it makes sense at a point is a question of
convergence.
The prototype: the geometric series
Take a = 0 and every coefficient c_n = 1.
The power series is the
geometric series
in x:
\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots.
Step 1 — the partial sum has a closed form. Multiply
S_N = 1 + x + \cdots + x^N by (1-x);
the middle terms telescope:
(1-x)S_N = 1 - x^{N+1} \quad\Longrightarrow\quad S_N = \frac{1 - x^{N+1}}{1 - x} \quad (x \ne 1).
Step 2 — take the limit. If |x| < 1 then
x^{N+1} \to 0, so S_N \to 1/(1-x). If
|x| \ge 1 the terms x^n do not even tend
to zero and the series diverges. Hence
\sum_{n=0}^{\infty} x^n = \frac{1}{1 - x}, \qquad \text{valid exactly on } (-1, 1).
A simple-looking sum on the left equals a genuine, named function on the right — but only
inside (-1,1). Where a power series converges is not optional; it
is part of what the function is.
Calculus, term by term
The real magic is that, inside its interval of convergence, a power series may be
differentiated and integrated one term at a time, just like a polynomial — and the new series
has the same interval. Differentiate the geometric series:
\frac{d}{dx}\sum_{n=0}^{\infty} x^n = \sum_{n=1}^{\infty} n x^{n-1} = \frac{d}{dx}\,\frac{1}{1-x} = \frac{1}{(1-x)^2}.
We never touched the limit machinery — we differentiated the closed form on the right and the
series on the left, and they agree. Likewise integrating term by term gives a new power series
for -\ln(1-x):
\int_0^x \frac{dt}{1-t} = \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} = x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots = -\ln(1-x).
One geometric series, differentiated and integrated, has just handed us power series for
1/(1-x)^2 and \ln. That is the engine of
the whole subject.
A power series f(x) = \sum_{n=0}^{\infty} c_n (x-a)^n defines a
function on the set of x where it converges. On the open interval
where it converges:
-
f is continuous, and indeed
infinitely differentiable.
-
Term-by-term differentiation is valid:
f'(x) = \sum_{n=1}^{\infty} n\, c_n (x-a)^{n-1}, with the same
interval of convergence.
-
Term-by-term integration is valid:
\int f = C + \sum_{n=0}^{\infty} \dfrac{c_n}{n+1}(x-a)^{n+1},
again with the same interval.
-
The geometric series is the prototype:
\sum_{n=0}^{\infty} x^n = \dfrac{1}{1-x} on
(-1,1).
A single convergent power series is automatically C^\infty on the
interior of its interval — you can differentiate it forever and every derivative is again a
power series. This is a small miracle. An arbitrary function built by other means might be
differentiable twice and then break; a power series never does. Each differentiation lowers
every exponent by one and multiplies by it, so
f^{(k)}(x) = \sum_{n=k}^{\infty} n(n-1)\cdots(n-k+1)\, c_n (x-a)^{n-k},
which is still a power series with the same radius. The catch — and it is a sharp one —
is the phrase "inside the interval". The convergence is what licenses every interchange of
limit, derivative and integral above. Step outside the interval and the series is not a
function at all, merely a divergent symbol. The next page,
radius of convergence,
is devoted to measuring exactly how wide that safe interval is.