Cauchy–Hadamard Root Test for Radius of Convergence

A power series $\sum_{n=0}^{\infty} c_n (z - a)^n$ converges on a disk centered at $a$; its radius of convergence $R$ is the half-width of that disk. The Cauchy–Hadamard theorem gives $R$ directly from the coefficients $c_n$, using the $\limsup$ so the formula is always valid — unlike the ratio test, which needs the limit of $|c_{n+1}/c_n|$ to exist.

The theorem

$\frac{1}{R} = \limsup_{n\to\infty} |c_n|^{1/n}, \qquad R = \frac{1}{\displaystyle\limsup_{n\to\infty} |c_n|^{1/n}}.$

Conventions: if the $\limsup$ is $0$ then $R = \infty$ (converges everywhere); if it is $\infty$ then $R = 0$ (converges only at the center). The series converges absolutely for $|z - a| < R$ and diverges for $|z - a| > R$; behavior exactly on $|z-a| = R$ must be checked separately.

How to apply

1. Identify the coefficients $c_n$ — the part multiplying $(z-a)^n$. If the powers are of some inner expression $u$, substitute $u$ first and solve for the radius in $u$.
2. Compute $L = \limsup_{n\to\infty} |c_n|^{1/n}$. When a true limit exists you may use $\lim$ in place of $\limsup$.
3. Invert: $R = 1/L$.
4. Undo any substitution. If $u = b\,x + d$, then $|u| < R$ becomes $|x + d/b| < R/|b|$, so the radius in $x$ is $R/|b|$ — the linear inner factor $b$ scales the radius by $1/|b|$.

Useful root-test limits: $n^{1/n} \to 1$, $(an)^{1/n} \to 1$, $a^{1/n} \to 1$ for $a > 0$, and $\left(1 - \tfrac1n\right)^{n} \to e^{-1}$.

Mini-example

Find $R$ for $\sum_{n=1}^{\infty} \dfrac{3^n}{n}\, x^n$. Here $c_n = \dfrac{3^n}{n}$, so

$|c_n|^{1/n} = \frac{3}{n^{1/n}} \longrightarrow \frac{3}{1} = 3 \implies R = \frac{1}{3}.$

The series converges for $|x| < \tfrac13$. (At the endpoints: $x = -\tfrac13$ gives the convergent alternating harmonic series, $x = \tfrac13$ the divergent harmonic series.)

Pitfall: when the variable sits inside a linear factor like $(b\,x + d)^n$, substitute $u = b\,x + d$, find the radius $R_u$ in $u$, then divide by $|b|$ to get the radius in $x$. Forgetting this factor is the classic distractor.