## Definition of limsup

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

${\displaystyle S_{N}=\{a_{N+1},a_{N+2},\ldots \},}$
the set obtained by throwing away the first ${\displaystyle N}$ terms in the sequence. Let
${\displaystyle M_{N}=\sup S_{N},}$
possibly equal to ${\displaystyle +\infty }$. Since ${\displaystyle S_{N}\subset S_{N-1}}$, it follows that ${\displaystyle M_{N}\leqslant M_{N-1}}$ for all ${\displaystyle N}$, so we have a non-increasing sequence of real numbers unless ${\displaystyle M_{N}=+\infty }$ for all ${\displaystyle N}$.

ThenDefinition (8):

With the above definitions,

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

where ${\displaystyle +\infty }$ and ${\displaystyle -\infty }$ are both possibilities.

Example: (9)

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

Example: (10)

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

${\displaystyle S_{2N-1}=\{1+1/(2N),-1,1+1/(2N+2),\ldots \}}$
The ${\displaystyle \sup }$ of this set is ${\displaystyle 1+1/2N}$. The limit as ${\displaystyle N\to +\infty }$ is ${\displaystyle 1}$.

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

Lemma (11):

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

${\displaystyle \limsup _{n\to \infty }a_{n}=L}$
Then given any ${\displaystyle \varepsilon >0}$ there exists ${\displaystyle N>0}$ such that which

${\displaystyle n>N_{0}\Longrightarrow a_{n}

Proof:

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

${\displaystyle N>N_{0}\Longrightarrow |M_{N}-L|<\varepsilon .}$
For such ${\displaystyle N}$,
${\displaystyle L-\varepsilon <\sup S_{N}
and so ${\displaystyle a_{n} for all ${\displaystyle n>N_{0}}$ as required.

${\displaystyle \Box }$

Remark:

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

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


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

Proof:

Put

${\displaystyle {\frac {1}{\rho }}=\limsup _{n\to \infty }|a_{n}|^{1/n}}$
Assume that ${\displaystyle \rho \neq 0,\infty }$. Suppose first that ${\displaystyle s, so
${\displaystyle |a_{n}|s^{n}\leqslant M(s)<\infty .}$
Taking ${\displaystyle n}$th roots,
${\displaystyle |a_{n}|^{1/n}\leqslant {\frac {M(s)^{1/n}}{s}}.}$
For any positive number ${\displaystyle C}$, ${\displaystyle \lim _{n\to \infty }C^{1/n}\to 1}$ as ${\displaystyle n\to \infty }$. By definition of the limit, it follows that for any given ${\displaystyle \varepsilon >0}$, there exists ${\displaystyle N}$ such that ${\displaystyle n>N}$ implies ${\displaystyle C^{1/n}<1+\varepsilon }$. Thus for ${\displaystyle n>N}$, ${\displaystyle |a_{n}|^{1/n}<(1+\varepsilon )/s}$, and so
${\displaystyle \limsup _{n\to \infty }|a_{n}|^{1/n}\leqslant {\frac {1+\varepsilon }{s}}.}$
This being true for every ${\displaystyle \varepsilon }$, we have Hence ${\displaystyle 1/\rho \leqslant 1/s}$ and ${\displaystyle s\leqslant \rho }$. This is true for every ${\displaystyle s, the radius of convergence, so finally
${\displaystyle r\leqslant \rho .}$
We aim to prove the opposite inequality. Suppose ${\displaystyle \varepsilon >0}$ is much smaller than ${\displaystyle \rho }$. Then
${\displaystyle {\frac {1}{\rho }}<{\frac {1}{\rho -\varepsilon }}}$
By the Lemma, we know that there exists ${\displaystyle N_{0}}$ so that ${\displaystyle n>N_{0}}$ implies that
${\displaystyle |a_{n}|^{1/n}>{\frac {1}{\rho -\varepsilon }}.}$
If ${\displaystyle s>\rho -\varepsilon }$, this implies
${\displaystyle |a_{n}|s^{n}>(\rho /(\rho -\varepsilon ))^{n}\to +\infty {\mbox{ as }}n\to \infty .}$
This means that ${\displaystyle M(s)=+\infty }$ for all ${\displaystyle s>\rho -\varepsilon }$, and any ${\displaystyle \varepsilon >0}$. This means that ${\displaystyle r\geqslant \rho }$. Combined with the previous inequality, we find that ${\displaystyle r=\rho }$. The cases ${\displaystyle \rho =0}$ and ${\displaystyle \rho =\infty }$ require slight modifications of the argument, and are left as exercises.

${\displaystyle \Box }$

Remark:

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

Proposition (14):

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

convergence ${\displaystyle r}$.

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

${\displaystyle \sum na_{n}z^{n-1}}$
is the same as that of
${\displaystyle \sum na_{n}z^{n}.}$
Now the ${\displaystyle n}$th root of the coefficient of ${\displaystyle z^{n}}$ in Definition of limsup is ${\displaystyle n^{1/n}|a_{n}|^{1/n}.}$Since ${\displaystyle \lim _{n\to \infty }n^{1/n}=1}$, it follows that
${\displaystyle \limsup _{n\to \infty }n^{1/n}|a_{n}|^{1/n}=\limsup _{n\to \infty }|a_{n}|^{1/n}=1/r.}$
By Hadamard's formula, we conclude that the radius of convergence of Definition of limsup is ${\displaystyle r}$.

${\displaystyle \Box }$