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

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

the set obtained by throwing away the first

$N$ terms in the
sequence. Let

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

possibly equal to

$+\infty$. Since

$S_{N}\subset S_{N-1}$, it follows
that

$M_{N}\leqslant M_{N-1}$ for all

$N$, so we have a non-increasing
sequence of real numbers unless

$M_{N}=+\infty$ for all

$N$.

Then

**Definition 8**

With the above definitions,

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

where

$+\infty$ and

$-\infty$ are both possibilities.

**Example 9**

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

**Example 10**

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

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

The

$\sup$ of this set is

$1+1/2N$. The limit as

$N\to +\infty$ is

$1$.

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

**Lemma 11**

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

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

Then given any

$\varepsilon >0$ there exists

$N>0$ such that
which

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

*Proof *

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

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

For such

$N$,

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

and so

$a_{n}<L+\varepsilon$ for all

$n>N_{0}$ as required.

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

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

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

for the radius of convergence.

*Proof *

Put

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

Assume that

$\rho \neq 0,\infty$.
Suppose first that

$s<r$, so

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

Taking

$n$th roots,

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

For any positive number

$C$,

$\lim _{n\to \infty }C^{1/n}\to 1$ as

$n\to \infty$.
By definition of the limit, it follows that for any given

$\varepsilon >0$,
there exists

$N$ such that

$n>N$ implies

$C^{1/n}<1+\varepsilon$. Thus
for

$n>N$,

$|a_{n}|^{1/n}<(1+\varepsilon )/s$, and so

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

This being true for every

$\varepsilon$, we have
Hence

$1/\rho \leqslant 1/s$ and

$s\leqslant \rho$. This is true for every

$s<r$, the radius of convergence, so finally

$r\leqslant \rho .$

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

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

By the
Lemma, we know that there exists

$N_{0}$ so that

$n>N_{0}$ implies that

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

If

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

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

This means that

$M(s)=+\infty$ for all

$s>\rho -\varepsilon$, and any

$\varepsilon >0$. This means that

$r\geqslant \rho$. Combined with the previous
inequality, we find that

$r=\rho$. The cases

$\rho =0$ and

$\rho =\infty$ require slight modifications of the argument, and are
left as exercises.

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

**Proposition 14**

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

*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

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

is the same as that of

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

Now the

$n$th root of the coefficient of

$z^{n}$ in

Definition of limsup
is

$n^{1/n}|a_{n}|^{1/n}.$Since

$\lim _{n\to \infty }n^{1/n}=1$, it
follows that

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

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

Definition of limsup is $r$.