Differentiate and Integrate Term by Term

Why uniform convergence is the engine

Swapping an infinite sum with a derivative or an integral is not automatic — it needs uniform convergence, not just pointwise. The good news for power series is that they converge uniformly on every closed interval strictly inside the radius. The tool is the Weierstrass M-test.

Claim: sum c_n x^n converges uniformly on [-r, r] for any 0 < r < R.

Pick rho with r < rho < R.  Since rho < R the series converges at rho,
so its terms are bounded: |c_n| rho^n <= M for some M and all n.

For every x in [-r, r]:
   |c_n x^n| <= |c_n| r^n = (|c_n| rho^n) (r/rho)^n <= M (r/rho)^n.

Let M_n = M (r/rho)^n.  Since r/rho < 1, sum M_n is a convergent
geometric series, and |c_n x^n| <= M_n for ALL x in [-r, r].

By the Weierstrass M-test, sum c_n x^n converges UNIFORMLY on [-r, r].  QED

Term-by-term operations keep the same radius

Differentiating sum c_n x^n term by term gives sum n c_n x^(n-1); integrating gives sum c_n x^(n+1)/(n+1). The first crucial fact: both new series have the same radius R as the original. This follows from Cauchy–Hadamard because n^(1/n) -> 1, so multiplying coefficients by n or by 1/(n+1) does not change the limsup of the n-th roots.

The second crucial fact is that these new series actually represent the derivative and antiderivative of the original sum. Call f(x) = sum c_n x^n on (-R, R). Then term-by-term differentiation gives f'(x) = sum n c_n x^(n-1), and term-by-term integration gives the integral from 0 to x of f equal to sum c_n x^(n+1)/(n+1). Both are legitimate inside the radius.

1. For integration: uniform convergence on [0, x] lets you interchange the integral and the infinite sum (uniform convergence preserves integrals). That gives the term-by-term integral directly.
2. For differentiation: the differentiated series sum n c_n x^(n-1) converges uniformly on closed subintervals (same M-test argument), and the original converges at one point. The theorem on differentiating a uniformly convergent series of derivatives then gives f' = that series.
3. Iterate: f' is again a power series with radius R, so it too is differentiable. By induction f has derivatives of every order — it is a smooth function on (-R, R).

A worked payoff

Start from the geometric series (|x| < 1):
   1/(1 - x) = sum_{n>=0} x^n.

Differentiate term by term (legal, same radius R = 1):
   1/(1 - x)^2 = sum_{n>=1} n x^(n-1) = 1 + 2x + 3x^2 + ...

Replace x by -x in the geometric series:
   1/(1 + x) = sum_{n>=0} (-1)^n x^n.

Integrate term by term from 0 to x (legal for |x| < 1):
   ln(1 + x) = sum_{n>=0} (-1)^n x^(n+1)/(n+1)
             = x - x^2/2 + x^3/3 - x^4/4 + ...

No new convergence work was needed: each operation kept R = 1.