Hadamard's n-th root formula

Definition of limsup

Let $(a_{n})$$(a_{n})$ be a sequence of real numbers. For each $N$$N$, define

$$S_{N}=\{a_{N+1},a_{N+2},\ldots \},$$

$$S_{N}=\{a_{N+1},a_{N+2},\ldots \},$$

the set obtained by throwing away the first $N$$N$ terms in the sequence. Let

$$M_{N}=\sup S_{N},$$

$$M_{N}=\sup S_{N},$$

possibly equal to $+\infty$$+\infty$. Since $S_{N}\subset S_{N-1}$$S_{N}\subset S_{N-1}$, it follows that $M_{N}\leqslant M_{N-1}$$M_{N}\leqslant M_{N-1}$ for all $N$$N$, so we have a non-increasing sequence of real numbers unless $M_{N}=+\infty$$M_{N}=+\infty$ for all $N$$N$.

Then

Definition 8

With the above definitions,

$$\limsup _{n\to \infty }a_{n}=\lim _{N\to \infty }M_{N}.$$

$$\limsup _{n\to \infty }a_{n}=\lim _{N\to \infty }M_{N}.$$

where $+\infty$$+\infty$ and $-\infty$$-\infty$ are both possibilities.

 

Example 9

If $a_{n}\to \ell$$a_{n}\to \ell$ as $n\to \infty$$n\to \infty$, then
 $\limsup _{n\to \infty }a_{n}=\ell$$\limsup _{n\to \infty }a_{n}=\ell$.

 

Example 10

Suppose $a_{n}=1+1/n$$a_{n}=1+1/n$ if $n$$n$ is even but $a_{n}=0$$a_{n}=0$ if $n$$n$ is odd. Then

$$S_{2N-1}=\{1+1/(2N),-1,1+1/(2N+2),\ldots \}$$

$$S_{2N-1}=\{1+1/(2N),-1,1+1/(2N+2),\ldots \}$$

The $\sup$$\sup$ of this set is $1+1/2N$$1+1/2N$. The limit as $N\to +\infty$$N\to +\infty$ is $1$$1$.

 

Intuitively, if $\limsup _{n\to \infty }a_{n}$$\limsup _{n\to \infty }a_{n}$ is finite, then it is the highest horizontal asymptote for the sequence (when graphed against $n$$n$).

Lemma 11

 Suppose that $(a_{n})$$(a_{n})$ is a real sequence with

$$\limsup _{n\to \infty }a_{n}=L$$

$$\limsup _{n\to \infty }a_{n}=L$$

Then given any $\varepsilon >0$$\varepsilon >0$ there exists $N>0$$N>0$ such that which

$$n>N_{0}\Longrightarrow a_{n}<L+\varepsilon .$$

$$n>N_{0}\Longrightarrow a_{n}<L+\varepsilon .$$

 

Proof

From the definition, there exists $N_{0}>0$$N_{0}>0$ so that

$$N>N_{0}\Longrightarrow |M_{N}-L|<\varepsilon .$$

$$N>N_{0}\Longrightarrow |M_{N}-L|<\varepsilon .$$

For such $N$$N$,

$$L-\varepsilon <\sup S_{N}<L+\varepsilon .$$

$$L-\varepsilon <\sup S_{N}<L+\varepsilon .$$

and so $a_{n}<L+\varepsilon$$a_{n}<L+\varepsilon$ for all $n>N_{0}$$n>N_{0}$ as required.

 

Remark 12

The first inequality in Definition of limsup shows that for each $N>N_{0}$$N>N_{0}$ there exists $n>N$$n>N$ such that $a_{n}>L-\varepsilon$$a_{n}>L-\varepsilon$. By taking an increasing sequence of $N$$N$, it follows that $a_{n}>L-\varepsilon$$a_{n}>L-\varepsilon$ for infinitely many $n$$n$.

 

We are now ready to prove Hadamard's 'formula' for the radius of convergence of a complex power series.

Theorem 4

For the power series  Radius of convergence , we have the formula

$${\frac {1}{r}}=\limsup _{n\to \infty }|a_{n}|^{1/n}$$

$${\frac {1}{r}}=\limsup _{n\to \infty }|a_{n}|^{1/n}$$

for the radius of convergence.

 

Proof

Put

$${\frac {1}{\rho }}=\limsup _{n\to \infty }|a_{n}|^{1/n}$$

$${\frac {1}{\rho }}=\limsup _{n\to \infty }|a_{n}|^{1/n}$$

Assume that $\rho \neq 0,\infty$$\rho \neq 0,\infty$. Suppose first that $s<r$$s<r$, so

$$|a_{n}|s^{n}\leqslant M(s)<\infty .$$

$$|a_{n}|s^{n}\leqslant M(s)<\infty .$$

Taking $n$$n$th roots,

$$|a_{n}|^{1/n}\leqslant {\frac {M(s)^{1/n}}{s}}.$$

$$|a_{n}|^{1/n}\leqslant {\frac {M(s)^{1/n}}{s}}.$$

For any positive number $C$$C$, $\lim _{n\to \infty }C^{1/n}\to 1$$\lim _{n\to \infty }C^{1/n}\to 1$ as $n\to \infty$$n\to \infty$. By definition of the limit, it follows that for any given $\varepsilon >0$$\varepsilon >0$, there exists $N$$N$ such that $n>N$$n>N$ implies $C^{1/n}<1+\varepsilon$$C^{1/n}<1+\varepsilon$. Thus for $n>N$$n>N$, $|a_{n}|^{1/n}<(1+\varepsilon )/s$$|a_{n}|^{1/n}<(1+\varepsilon )/s$, and so

$$\limsup _{n\to \infty }|a_{n}|^{1/n}\leqslant {\frac {1+\varepsilon }{s}}.$$

$$\limsup _{n\to \infty }|a_{n}|^{1/n}\leqslant {\frac {1+\varepsilon }{s}}.$$

This being true for every $\varepsilon$$\varepsilon$, we have Hence $1/\rho \leqslant 1/s$$1/\rho \leqslant 1/s$ and $s\leqslant \rho$$s\leqslant \rho$. This is true for every $s<r$$s<r$, the radius of convergence, so finally

$$r\leqslant \rho .$$

$$r\leqslant \rho .$$

We aim to prove the opposite inequality. Suppose $\varepsilon >0$$\varepsilon >0$ is much smaller than $\rho$$\rho$. Then

$${\frac {1}{\rho }}<{\frac {1}{\rho -\varepsilon }}$$

$${\frac {1}{\rho }}<{\frac {1}{\rho -\varepsilon }}$$

By the Lemma, we know that there exists $N_{0}$$N_0$ so that $n>N_{0}$$n>N_{0}$ implies that

$$|a_{n}|^{1/n}>{\frac {1}{\rho -\varepsilon }}.$$

$$|a_{n}|^{1/n}>{\frac {1}{\rho -\varepsilon }}.$$

If $s>\rho -\varepsilon$$s>\rho -\varepsilon$, this implies

$$|a_{n}|s^{n}>(\rho /(\rho -\varepsilon ))^{n}\to +\infty {\mbox{ as }}n\to \infty .$$

$$|a_{n}|s^{n}>(\rho /(\rho -\varepsilon ))^{n}\to +\infty {\mbox{ as }}n\to \infty .$$

This means that $M(s)=+\infty$$M(s)=+\infty$ for all $s>\rho -\varepsilon$$s>\rho -\varepsilon$, and any $\varepsilon >0$$\varepsilon >0$. This means that $r\geqslant \rho$$r\geqslant \rho$. Combined with the previous inequality, we find that $r=\rho$$r=\rho$. The cases $\rho =0$$\rho =0$ and $\rho =\infty$$\rho =\infty$ require slight modifications of the argument, and are left as exercises.

 

Remark 13

The $n$$n$th root test has the advantage that it works always, provided one can calculate the $\limsup$$\limsup$, of course. This in contrast to the ratio test, which only works if $\lim |a_{n+1}/a_{n}|$$\lim |a_{n+1}/a_{n}|$ exists.

 

Proposition 14

If $\sum _{n=0}^{\infty }a_{n}z^{n}$$\sum _{n=0}^{\infty }a_{n}z^{n}$ has radius of convergence $r$$r$, then the series $\sum _{n=0}^{\infty }na_{n}z^{n-1}$$\sum _{n=0}^{\infty }na_{n}z^{n-1}$ has the same radius of convergence $r$$r$.

 

Proof

This can be proved by more elementary

 means, but 

we shall prove it as an illustration of Hadamard's formula.

We have observed in Remark Radius of convergence

 that the radius of convergence of

$$\sum na_{n}z^{n-1}$$

$$\sum na_{n}z^{n-1}$$

is the same as that of

$$\sum na_{n}z^{n}.$$

$$\sum na_{n}z^{n}.$$

Now the $n$$n$th root of the coefficient of $z^{n}$$z^{n}$ in Definition of limsup is $n^{1/n}|a_{n}|^{1/n}.$$n^{1/n}|a_{n}|^{1/n}.$Since $\lim _{n\to \infty }n^{1/n}=1$$\lim _{n\to \infty }n^{1/n}=1$, it follows that

$$\limsup _{n\to \infty }n^{1/n}|a_{n}|^{1/n}=\limsup _{n\to \infty }|a_{n}|^{1/n}=1/r.$$

$$\limsup _{n\to \infty }n^{1/n}|a_{n}|^{1/n}=\limsup _{n\to \infty }|a_{n}|^{1/n}=1/r.$$

By Hadamard's formula, we conclude that the radius of convergence of

Definition of limsup  is $r$$r$.