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Cauchy–Hadamard: Computing the Radius

One formula gives R for every power series: it reads the radius off the limit superior of the n-th roots of the coefficients. We prove it, see why the ratio test is a friendlier special case, and check the endpoints by hand.

The formula and why it must be true

The Cauchy–Hadamard theorem says the radius R of sum c_n (x - a)^n is given by 1/R = limsup of |c_n|^(1/n) as n -> infinity, with the conventions 1/0 = infinity and 1/infinity = 0. The reason is that the root test applied at a point x reads off precisely this quantity.

Apply the root test to the absolute series sum |c_n| |x - a|^n.

Let L = limsup |c_n|^(1/n).  The n-th root of the n-th term is

   ( |c_n| |x - a|^n )^(1/n) = |c_n|^(1/n) * |x - a|.

Take limsup over n  ->  L * |x - a|.

Root test:
   if L * |x - a| < 1  the series converges absolutely;
   if L * |x - a| > 1  the n-th term does not -> 0, so it diverges.

The threshold is L * |x - a| = 1, i.e. |x - a| = 1/L.
That threshold is exactly the radius:  R = 1/L = 1 / limsup |c_n|^(1/n).
The root test turns the radius into a one-line computation.

The ratio test, an easier special case

In practice the ratio test is often easier. When the limit lim |c_{n+1} / c_n| exists, it equals limsup |c_n|^(1/n), so you may use R = lim |c_n / c_{n+1}|. The ratio test is less general than the root test — it can fail to give a limit even when Cauchy–Hadamard works — but for factorials and clean closed-form coefficients it is the fast route.

Example: R for sum x^n / n!  (coefficients c_n = 1/n!).

Ratio test on |c_{n+1} / c_n|:
   |c_{n+1} / c_n| = (1/(n+1)!) / (1/n!) = n! / (n+1)! = 1/(n+1) -> 0.

So 1/R = 0, hence R = infinity.  Converges for all x.    (the exp series)

Example: R for sum n! x^n  (c_n = n!).
   |c_{n+1} / c_n| = (n+1)!/n! = n+1 -> infinity.
So 1/R = infinity, hence R = 0.  Converges only at x = 0.

Example: R for sum x^n / n  (c_n = 1/n).
   |c_{n+1}/c_n| = n/(n+1) -> 1.  So R = 1.
Three radii from the ratio of consecutive coefficients.

Endpoints decide the interval

Cauchy–Hadamard gives R but says nothing at |x - a| = R. The full interval of convergence is found by testing each endpoint separately. The series sum x^n / n with R = 1 is the classic illustration that the two ends can disagree.

Series sum_{n>=1} x^n / n,  radius R = 1, center 0.

Right endpoint x = 1:  sum 1/n = harmonic series -> DIVERGES.

Left endpoint x = -1:  sum (-1)^n / n.
   This is alternating; terms 1/n decrease monotonically to 0.
   By the alternating series test -> CONVERGES (conditionally).

Conclusion: interval of convergence is [-1, 1),
   closed on the left, open on the right.  The two ends differ.
Same radius, opposite endpoint behavior — always test both.