Vittorio Reggianini (1858-1939) was an Italian painter.
The concepts of infimum and supremum are needed for studying the limit superior and limit inferior. You may refer to the page infimum and supremum for more details.
Let \(\{a_{n}\}_{n=1}^{\infty}\) be a sequence of real numbers. When the limit of sequence \(\{a_{n}\}_{n=1}^{\infty}\) does not be convergent, we are going to introduce the weak limit concepts called limit superior and limit inferior.
- We say that the sequence is increasing when \(a_{n}\leq a_{n+1}\) for \(n=1,2,\cdots\).
- We say that the sequence is decreasing when \(a_{n}\geq a_{n+1}\) for \(n=1,2,\cdots\).
The sequence \(\{a_{n}\}_{n=1}^{\infty}\) is called monotonic when it is increasing or decreasing.
Proposition. A monotonic sequence is convergent if and only if it is bounded.
Proof. Suppose that \(\{a_{n}\}_{n=1}^{\infty}\) is increasing. Then, we have
\[\lim_{n\rightarrow\infty}a_{n}=\sup_{n}a_{n}.\]
Suppose that \(\{a_{n}\}_{n=1}^{\infty}\) is decreasing. Then, we have
\[\lim_{n\rightarrow\infty}a_{n}=\inf_{n}a_{n}.\]
This completes the proof. \(\blacksquare\)
Definition. Let \(\{a_{n}\}_{n=1}^{\infty}\) be a sequence of real numbers. The limit superior of \(\{a_{n}\}_{n=1}^{\infty}\) is denoted and defined by
\[\limsup_{n\rightarrow\infty}a_{n}=\inf_{n\geq 1}\sup_{k\geq n}a_{k}\]
and the limit inferior of \(\{a_{n}\}_{n=1}^{\infty}\) is denoted and defined by
\[\liminf_{n\rightarrow\infty}a_{n}=\sup_{n\geq 1}\inf_{k\geq n}a_{k}.\]
It is clear to see
\[\liminf_{n\rightarrow\infty}a_{n}=-\limsup_{n\rightarrow\infty}(-a_{n}).\]
For each \(n\), let
\begin{equation}{\label{ma9}}\tag{1}b_{n}=\sup_{k\geq n}a_{k}\mbox{ and }c_{n}=\inf_{k\geq n}a_{k}\end{equation}
We see that \(\{b_{n}\}_{n=1}^{\infty}\) is a decreasing sequence and \(\{c_{n}\}_{n=1}^{\infty}\) is an increasing sequence. In this case, we have
\[\inf_{n\geq 1}b_{n}=\lim_{n\rightarrow\infty}b_{n}\mbox{ and }
\sup_{n\geq 1}c_{n}=\lim_{n\rightarrow\infty}c_{n},\]
which also says
\begin{equation}{\label{ma10}}\tag{2}\limsup_{n\rightarrow\infty}a_{n}=\inf_{n\geq 1}b_{n}=\lim_{n\rightarrow\infty}b_{n}=\lim_{n\rightarrow\infty}\sup_{k\geq n}a_{k}\end{equation}
and
\begin{equation}{\label{ma11}}\tag{3}\liminf_{n\rightarrow\infty}a_{n}=\sup_{n\geq 1}c_{n}=\lim_{n\rightarrow\infty}c_{n}=\lim_{n\rightarrow\infty}\inf_{k\geq n}a_{k}.\end{equation}
Proposition. Let \(\{a_{n}\}_{n=1}^{\infty}\) be a sequence of real numbers. We have the following results.
(i) We have
\[\liminf_{n\rightarrow\infty}a_{n}\leq\limsup_{n\rightarrow\infty}a_{n}.\]
(ii) We have that
\[\lim_{n\rightarrow\infty}a_{n}=a\mbox{ if and only if }\liminf_{n\rightarrow\infty}a_{n}=\limsup_{n\rightarrow\infty}a_{n}=a
\mbox{ with }|a|<+\infty .\]
(iii) The sequence diverges to \(+\infty\) if and only if
\[\liminf_{n\rightarrow\infty}a_{n}=\limsup_{n\rightarrow\infty}a_{n}=+\infty .\]
(iv) The sequence diverges to \(-\infty\) if and only if
\[\liminf_{n\rightarrow\infty}a_{n}=\limsup_{n\rightarrow\infty}a_{n}=-\infty.\]
(v) Let \(\{b_{n}\}_{n=1}^{\infty}\) be another sequence satisfying \(a_{n}\leq b_{n}\) for each \(n=1,2,\cdots\). Then, we have
\[\liminf_{n\rightarrow\infty}a_{n}\leq\liminf_{n\rightarrow\infty}b_{n}\mbox{ and }
\limsup_{n\rightarrow\infty}a_{n}\leq\limsup_{n\rightarrow\infty}b_{n}.\]
Proof. To prove part (i), from (\ref{ma9}), we see that \(c_{n}\leq b_{n}\) for all \(n\geq 1\). Using (\ref{ma10}) and (\ref{ma11}), we obtain
\[\liminf_{n\rightarrow\infty}a_{n}=\lim_{n\rightarrow\infty}c_{n}\leq
\lim_{n\rightarrow\infty}b_{n}=\limsup_{n\rightarrow\infty}a_{n}.\]
To prove part (ii), suppose that
\[\lim_{n\rightarrow\infty}a_{n}=a.\]
Then, given any \(\epsilon >0\), there exists an integer \(N\) satisfying
\[a-\frac{\epsilon}{2}<a_{n}<a+\frac{\epsilon}{2}\mbox{ for }n\geq N,\]
which implies
\[a-\frac{\epsilon}{2}\leq\inf_{k\geq n}a_{k}=c_{n}\mbox{ and }b_{n}=\sup_{k\geq n}a_{k}\leq
a+\frac{\epsilon}{2}\mbox{ for }n\geq N.\]
In other words, we have
\[a-\frac{\epsilon}{2}\leq c_{n}\leq b_{n}\leq a+\frac{\epsilon}{2}\mbox{ for }n\geq N,\]
which also implies
\[\left |c_{n}-a\right |\leq\frac{\epsilon}{2}<\epsilon\mbox{ and }
\left |b_{n}-a\right |\leq\frac{\epsilon}{2}<\epsilon\mbox{ for }n\geq N.\]
Therefore, we obtain
\[\lim_{n\rightarrow\infty}c_{n}=a=\lim_{n\rightarrow\infty}b_{n},\]
which implies, by using (\ref{ma10}) and (\ref{ma11}),
\[\liminf_{n\rightarrow\infty}a_{n}=\limsup_{n\rightarrow\infty}a_{n}=a.\]
For the converse, from (\ref{ma9}) again, we see that \(c_{n}\leq a_{n}\leq b_{n}\) for all \(n\geq 1\). Since
\[a=\lim_{n\rightarrow\infty}\inf_{k\geq n}a_{k}=\lim_{n\rightarrow\infty}c_{n}\mbox{ and }
a=\lim_{n\rightarrow\infty}\sup_{k\geq n}a_{k}=\lim_{n\rightarrow\infty}b_{n},\]
using the pinching theorem, we obtain the desired result. The other parts are not difficult to obtain. This completes the proof. \(\blacksquare\)
Proposition. Let \(\{a_{n}\}_{n=1}^{\infty}\) and \(\{b_{n}\}_{n=1}^{\infty}\) be any two sequences. Then, we have
\[\limsup_{n\rightarrow\infty}\left (a_{n}+b_{n}\right )\leq
\limsup_{n\rightarrow\infty}a_{n}+\limsup_{n\rightarrow\infty}b_{n}\]
and
\[\liminf_{n\rightarrow\infty}\left (a_{n}+b_{n}\right )\geq
\liminf_{n\rightarrow\infty}a_{n}+\liminf_{n\rightarrow\infty}b_{n}.\]
Proof. For \(k\geq n\), we have
\[a_{k}+b_{k}\leq\sup_{k\geq n}a_{k}+\sup_{k\geq n}b_{k},\]
which says that
\begin{equation}{\label{ma12}}\tag{4}\sup_{k\geq n}\left (a_{k}+b_{k}\right )\leq\sup_{k\geq n}a_{k}+\sup_{k\geq n}b_{k}.\end{equation}
Therefore, we obtain
\begin{align*}\limsup_{n\rightarrow\infty}\left (a_{n}+b_{n}\right ) & =\inf_{n\geq 1}\sup_{k\geq n}\left (a_{k}+b_{k}\right )=\lim_{n\rightarrow\infty}\sup_{k\geq n}\left (a_{k}+b_{k}\right )\\& \leq\lim_{n\rightarrow\infty}\left [\sup_{k\geq n}a_{k}+\sup_{k\geq n}b_{k}\right ]\mbox{ (using (\ref{ma12}))}\\ & =\lim_{n\rightarrow\infty}\sup_{k\geq n}a_{k}+\lim_{n\rightarrow\infty}\sup_{k\geq n}b_{k}\mbox{ (since the limits exist)}\\ & =\limsup_{n\rightarrow\infty}a_{n}+\liminf_{n\rightarrow\infty}\mbox{ (using (\ref{ma10}) and (\ref{ma11}))}.\end{align*}
Now, we similarly have
\begin{equation}{\label{ma13}}\tag{5}\inf_{k\geq n}\left (a_{k}+b_{k}\right )\geq\inf_{k\geq n}a_{k}+\inf_{k\geq n}b_{k}. \end{equation}
Therefore, we also obtain
\begin{align*} \liminf_{n\rightarrow\infty}\left (a_{n}+b_{n}\right ) & =\sup_{n\geq 1}\inf_{k\geq n}\left (a_{k}+b_{k}\right ) =\lim_{n\rightarrow\infty}\inf_{k\geq n}\left (a_{k}+b_{k}\right )\\ & \geq\lim_{n\rightarrow\infty}\left [\inf_{k\geq n}a_{k}+\inf_{k\geq n}b_{k}\right ] \mbox{ (using (\ref{ma13}))}\\ & =\lim_{n\rightarrow\infty}\inf_{k\geq n}a_{k}+\lim_{n\rightarrow\infty}\inf_{k\geq n}b_{k} \mbox{ (since the limits exist)}\\ & =\liminf_{n\rightarrow\infty}a_{n}+\liminf_{n\rightarrow\infty}b_{n} \mbox{ (using (\ref{ma10}) and (\ref{ma11}))}. \end{align*}
This completes the proof. \(\blacksquare\)
Example. Given a sequence \(\{a_{n}\}_{n=1}^{\infty}\) by
\[a_{n}=(-1)^{n}\left (1+\frac{1}{n}\right ),\]
We have
\[b_{n}=\sup_{k\geq n}a_{k}=\sup_{k\geq n}(-1)^{k}\left (1+\frac{1}{k}\right )=\left\{\begin{array}{ll}1+\frac{1}{n} & \mbox{if \(n\) is even}\\ 1+\frac{1}{n+1} & \mbox{if \(n\) is odd}\end{array}\right .\]
and
\[c_{n}=\inf_{k\geq n}a_{k}=\inf_{k\geq n}(-1)^{k}\left (1+\frac{1}{k}\right )=\left\{\begin{array}{ll}-\left (1+\frac{1}{n+1}\right ) & \mbox{if \(n\) is even}\\-\left (1+\frac{1}{n}\right ) & \mbox{if \(n\) is odd}\end{array}\right .\]
Therefore, we obtain
\[\limsup_{n\rightarrow}a_{n}=\lim_{n\rightarrow\infty}b_{n}=0\]
and
\[\liminf_{n\rightarrow}a_{n}=\lim_{n\rightarrow\infty}c_{n}=-1.\]
This also says that the limit does not exist. \(\sharp\)
Example. Given a sequence \(\{a_{n}\}_{n=1}^{\infty}\) by
\[a_{n}=(-1)^{n}+\frac{1}{2^{n}},\]
we have
\[b_{n}=\sup_{k\geq n}a_{k}=\sup_{k\geq n}(-1)^{k}\left [(-1)^{n}+\frac{1}{2^{n}}\right ]=\left\{\begin{array}{ll}1+\frac{1}{2^{n}} & \mbox{if \(n\) is even}\\1+\frac{1}{2^{n+1}} & \mbox{if \(n\) is odd}\end{array}\right .\]
and
\[c_{n}=\inf_{k\geq n}a_{k}=\inf_{k\geq n}(-1)^{k}\left [(-1)^{n}+\frac{1}{2^{n}}\right ]=\left\{\begin{array}{ll}-1 & \mbox{if \(n\) is even}\\-1 & \mbox{if \(n\) is odd}\end{array}\right .\]
Therefore, we obtain
\[\limsup_{n\rightarrow}a_{n}=\lim_{n\rightarrow\infty}b_{n}=1\]
and
\[\liminf_{n\rightarrow}a_{n}=\lim_{n\rightarrow\infty}c_{n}=-1.\]
This also says that the limit does not exist. \(\sharp\)
Proposition. Given a positive sequence \(\{x_{n}\}_{n=1}^{\infty}\) with \(x_{n}>0\) for all \(n\), we have
\[\liminf_{n\rightarrow\infty}\frac{x_{n+1}}{x_{n}}\leq\liminf_{n\rightarrow\infty}\sqrt[n]{x_{n}}
\leq\limsup_{n\rightarrow\infty}\sqrt[n]{x_{n}}\leq
\limsup_{n\rightarrow\infty}\frac{x_{n+1}}{x_{n}}.\]
Suppose that the limit \(\lim_{n\rightarrow\infty}(x_{n+1}/x_{n})\) exists. Then the limit \(\lim_{n\rightarrow\infty}\sqrt[n]{x_{n}}\) also exists.
Proof. We have
\[L=\liminf_{n\rightarrow\infty}\frac{x_{n+1}}{x_{n}}=\sup_{n\geq 1}\inf_{k\geq n}
\frac{x_{n+1}}{x_{n}}=\lim_{n\rightarrow\infty}c_{n},\mbox{ where }c_{n}=\inf_{k\geq n}
\frac{x_{n+1}}{x_{n}}.\]
Therefore, given any \(\epsilon>0\), there exists an integer \(N\) satisfying
\[L-\epsilon<c_{n}=\inf_{k\geq n}\frac{x_{n+1}}{x_{n}}\mbox{ for }n\geq N,\]
which implies
\[(L-\epsilon)x_{n}\leq x_{n+1}\mbox{ for }n\geq N.\]
By induction, we can obtain
\[(L-\epsilon)^{n-N}x_{N}\leq x_{n}\mbox{ for }n\geq N,\]
which also implies
\[(L-\epsilon)^{(n-N)/n}\sqrt[n]{x_{N}}\leq\sqrt[n]{x_{n}}\mbox{ for }n\geq N.\]
Therefore, since \(x_{N}\) is a fixed number, we obtain
\[L-\epsilon=\lim_{n\rightarrow\infty}(L-\epsilon)^{(n-N)/n}\sqrt[n]{x_{N}}\leq
\lim_{n\rightarrow\infty}\sqrt[n]{x_{n}}.\]
Since \(\epsilon\) can be any positive numbers, it follows
\[\liminf_{n\rightarrow\infty}\frac{x_{n+1}}{x_{n}}=
L\leq\lim_{n\rightarrow\infty}\sqrt[n]{x_{n}}.\]
We can similarly show that
\[\limsup_{n\rightarrow\infty}\sqrt[n]{x_{n}}\leq\limsup_{n\rightarrow\infty}\frac{x_{n+1}}{x_{n}}.\]
When the limit \(\lim_{n\rightarrow\infty}(x_{n+1}/x_{n})\) exists, we have
\[\liminf_{n\rightarrow\infty}\frac{x_{n+1}}{x_{n}}=\limsup_{n\rightarrow\infty}\frac{x_{n+1}}{x_{n}},\]
which also implies
\[\liminf_{n\rightarrow\infty}\sqrt[n]{x_{n}}=\limsup_{n\rightarrow\infty}\sqrt[n]{x_{n}}.\]
This completes the proof. \(\blacksquare\)
