What is Radius of convergence?

Example 11.3 (A radius with no ratio test)

What is the radius of $\sum \sin(n)\,z^n$? The ratio $\abs{\sin(n+1)/\sin n}$ has no limit, but the definition works directly. $R \geq 1$: $\abs{\sin n} \leq 1$, so $(\sin n\cdot r^n)$ is bounded for every $r < 1$ — indeed for $r = 1$. $R \leq 1$: it suffices that $\sin n \not\to 0$. Suppose $\sin n \to 0$; the addition formula

would force $\cos n \to 0$ (solve for $\cos n$: $\sin 1 \neq 0$), contradicting $\sin^2 n + \cos^2 n = 1$. So the terms $\sin(n)\,1^n$ do not tend to $0$: the series diverges at $z = 1$, and $R \leq 1$. Conclusion: $R = 1$. Closing insight: the radius is a statement about $\abs{a_n}r^n$ being bounded — no limit of ratios is ever required, and boundedness arguments settle cases the ratio test cannot touch (compare the oscillating coefficients of Exercise 11.1).

Example 11.6 (Multiplying by $\frac{1}{1-x}$ sums the coefficients)

A Cauchy product against the geometric series has a memorable meaning: for any $\sum a_nx^n$ of radius $R > 0$ and $\abs x < \min(R, 1)$,

multiplication by $\frac{1}{1-x}$ replaces coefficients by their partial sums (convolve with the all-ones sequence). Instance: $\dfrac{\eu^x}{1-x} = \sum_n s_n x^n$ with $s_n = \sum_{k\leq n}\frac{1}{k!}$, the partial sums of $\eu$ — compare Exercise 11.11, where the same product with $\eu^{-x}$ encodes the derangement counts. Closing insight: operations on power series are operations on coefficient sequences in disguise (multiply by $\frac1{1-x}$: sum; multiply by $x$: shift; differentiate: multiply by $n$ and shift) — a dictionary the generating-functions chapter will read fluently.

Example 11.20 (Binet’s formula, carried out)

Let $\varphi = \frac{1+\sqrt5}{2}$ and $\psi = \frac{1-\sqrt5}{2}$, the roots of $X^2 = X + 1$; since $\varphi + \psi = 1$ and $\varphi\psi = -1$,

Partial fractions: seeking $\frac{x}{(1-\varphi x)(1-\psi x)} = \frac{A}{1 - \varphi x} + \frac{B}{1 - \psi x}$, the constant term gives $A + B = 0$ and the $x$-coefficient $-A\psi - B\varphi = 1$, so $A(\varphi - \psi) = 1$: $A = \frac{1}{\sqrt5} = -B$. Two geometric series later,

by uniqueness of coefficients (Theorem 11.7). Since $\abs\psi < 1$, the term $\frac{\psi^n}{\sqrt5}$ has absolute value $< \frac12$: $F_n$ is the nearest integer to $\frac{\varphi^n}{\sqrt5}$. Closing insight: the radius $\frac1\varphi$ of $F$ is the reciprocal of the dominant root — growth of coefficients and radius of convergence are the same information read in opposite directions.