Radius of convergence and lim sup? Confused about lim sup

Question

$\lim \sup |a_n|^{1/n}=1/R$ where R is the radius of convergence. I am confused about what the lim sup is, is it that for each $n \in \mathbb N$ we get the sup of those $|a_n|^{1/n}$ then we take the limit as $n\to \infty$ ?

Answer 1

Not exactly: for each $n$, you get $\;\sup\limits_{\,k\ge n}\Bigl(|a_k|^{\tfrac1k}\Bigr)$.

This is a non-increasing sequence in $\mathbf R^+\cup\{+\infty\}$, hence by the monotone convergence theorem, it has (finite or infinite) limit. Thus $$\limsup_n\Bigl(|a_n|^{\tfrac1n}\Bigr)=\lim_{n}\Bigl(\sup_{k\ge n}\Bigl(|a_k|^{\tfrac1k}\Bigr)\Bigr).$$

Answer 2

What you are after is the concept of the limit superior of a sequence $(x_{n})$ of real numbers. Intuitive illustration may be helpful. To fix it, let us do the following operation: Take the supremum of $x_{n}$ over $n$ and denote $y_{1} := \sup_{n \geq 1}x_{n}$; delete $x_{1}$ and take $y_{2} := \sup_{n \geq 2}x_{n}$; having taking $y_{j} := \sup_{n \geq j}x_{n}$, take $y_{j+1} := \sup_{n \geq j+1}x_{n}$. Then $(y_{n})$ is a decreasing (why? :)) sequence so that $\lim_{n}y_{n}$ exists and $\inf_{n}y_{n} = \lim_{n}y_{n}$ (why? :)). Write things out to see that $\inf_{n}y_{n} = \inf_{n}\sup_{j \geq n}x_{n}$ by construction.

If you accept the construction above, then you get the concept of limit inferior of a sequence too!

Answer 3

The superior limit of a sequence $(a_n)_{n\in\mathbb N}$ (denoted by $\limsup_na_n$) is the greatest element of $\mathbb{R}\cup\{\pm\infty\}$ which is the limit of a subsequence of the sequence $(a_n)_{n\in\mathbb N}$. If the sequence has a limit $L\in\mathbb{R}\cup\{\pm\infty\}$, then $\limsup_na_n=L$.

Answer 4

Let $(a_n)$ be a bounded sequence. The sequence $y_n$ be defined by $y_n = sup $ {$a_k : n \le k$ } . Then limit superior of $a_n$ is defined as $ lim$ $sup $ $a_n = lim$ $ y_n $ .