Stolz theorem. Let \((x_n)_{n \in \mathbb{N}}\), \((y_n)_{n \in \mathbb{N}}\) be two sequences so that
\((y_n)_{n \in \mathbb{N}}\) strictly increases to \(+\infty\), i.e. \(y_n<y_{n+1}\) for all \(n \in \mathbb{N}\) and \[\lim\limits_{n \to \infty}y_n=+\infty.\]
Also we have \[\lim_{n \to \infty}\frac{x_n-x_{n-1}}{y_n-y_{n-1}}=a,\]
then \[\lim_{n \to \infty}\frac{x_n}{y_n}=a.\]
Remark: It is a prototype of a l’Hôpital’s rule.
Without loss of generality we may assume that \(y_{n} > 0\), since \(\lim\limits_{n \to \infty}y_n=+\infty\) and thus we have \(y_{n} > 0\) for large \(n \in \mathbb{N}\).
Since \(\lim\limits_{n \to \infty}\frac{x_n-x_{n-1}}{y_n-y_{n-1}}=a\), there is \(M>0\) such that for \(n \ge M\) we have \[\begin{aligned} a - \frac{\varepsilon}{2} < \frac{x_n-x_{n-1}}{y_n-y_{n-1}} < a + \frac{\varepsilon}{2} . \end{aligned}\] So for \(n \ge M\) we have \[\begin{aligned} {\color{blue} \Big(a -\frac{\varepsilon}{2} \Big) (y_n-y_{n-1}) < } x_n-x_{n-1} {\color{purple} < \Big(a + \frac{\varepsilon}{2} \Big) (y_n-y_{n-1}) }. \end{aligned}\]
Summing now from \(k=M\) to \(k=n\) for any \(n \ge M\) we get \[\begin{aligned} \Big(a - \frac{\varepsilon}{2} \Big) (y_{n}-y_{M-1}) & = {\color{blue} \sum_{k=M}^{n} \Big(a - \frac{\varepsilon}{2} \Big) (y_k-y_{k-1}) < } \sum_{k=M}^{n} (x_k-x_{k-1}) \\ = x_{n}-x_{M-1} & {\color{purple} < \sum_{k=M}^{n} \Big(a + \frac{\varepsilon}{2} \Big) (y_{k}-y_{k-1})} = \Big(a + \frac{\varepsilon}{2} \Big) (y_{n}-y_{M-1}). \end{aligned}\] since the above sums are telescoping.
Therefore dividing by \(y_{n}-y_{M-1}\) for any \(n \ge M\) we obtain \[\begin{aligned} a - \frac{\varepsilon}{2} < \frac{x_{n}-x_{M-1}}{y_{n}-y_{M-1}} < a + \frac{\varepsilon}{2} . \end{aligned}\] So \[\begin{aligned} {\color{blue} \Big| \frac{x_{n}-x_{M-1}}{y_{n}-y_{M-1}} - a \Big| < \frac{\varepsilon}{2} } \qquad \text{for $n \ge M$.} \end{aligned}\]
Observe that \[\begin{aligned} {\color{red} \Big| \frac{x_{n} }{y_{n}} - \Big( 1 - \frac{y_{M-1}}{y_{n}} \Big) \frac{x_{n}-x_{M-1}}{y_{n}-y_{M-1}} \Big| } = \Big| \frac{x_{n} }{y_{n}} - \frac{x_{n}-x_{M-1}}{y_{n}} \Big| = {\color{red} \Big| \frac{x_{M-1}}{y_{n}} \Big| }, \end{aligned}\] so the triangle inequality gives us for \(n \ge M\) the inequalities \[\begin{aligned} \Big| \frac{x_{n} }{y_{n}} - a \Big| & \le {\color{red} \Big| \frac{x_{n} }{y_{n}} - \Big( 1 - \frac{y_{M-1}}{y_{n}} \Big) \frac{x_{n}-x_{M-1}}{y_{n}-y_{M-1}} \Big| } + \Big| \Big( 1 - \frac{y_{M-1}}{y_{n}} \Big) \frac{x_{n}-x_{M-1}}{y_{n}-y_{M-1}} - a \Big| \\ & \le {\color{red} \Big| \frac{x_{M-1}}{y_{n}} \Big| } + \Big| \Big( 1 - \frac{y_{M-1}}{y_{n}} \Big) \Big| {\color{blue} \Big| \frac{x_{n}-x_{M-1}}{y_{n}-y_{M-1}} - a \Big| } + |a| \frac{y_{M-1}}{y_{n}} \\ & < \frac{\varepsilon}{2} + \frac{|x_{M-1}| + |a| y_{M-1}}{y_{n}}. \end{aligned}\]
So for \(n \ge M\) we have \[\begin{aligned} \Big| \frac{x_{n} }{y_{n}} - a \Big| < \frac{\varepsilon}{2} + \frac{|x_{M-1}| + |a| y_{M-1}}{y_{n}}. \end{aligned}\] Since \(\lim\limits_{n \to \infty}y_n=+\infty\), we may choose \(N\ge M\) such that for \(n \ge N\) we have \[\frac{|x_{M-1}| + |a| y_{M-1}}{y_{n}} < \frac{\varepsilon}{2}.\] Therefore for \(n \ge N\) we get \[\begin{aligned} \Big| \frac{x_{n} }{y_{n}} - a \Big| < \varepsilon \end{aligned}\] and the proof is finished. $$\tag*{$\blacksquare$}$$
Exercise. Let \(k \in \mathbb{N}\) be fixed. Find the limit \[\begin{aligned} \lim_{n \to \infty} \frac{1^k + \ldots + n^{k}}{n^{k+1}}. \end{aligned}\]
Proof. We apply Stolz’s theorem with \[\begin{aligned} x_{n} = 1^k + \ldots + n^{k}, \qquad y_n = n^{k+1}. \end{aligned}\] Observe that \((y_n)_{n \in \mathbb{N}}\) is strictly increasing and \(\lim\limits_{n \to \infty} y_{n} = + \infty\). Therefore it suffices to compute \[\begin{aligned} \lim_{n \to \infty} \frac{x_n-x_{n-1}}{y_n-y_{n-1}} = \lim_{n \to \infty} \frac{n^{k}}{n^{k+1} - (n-1)^{k+1}}. \end{aligned}\]
Using the binomial theorem we get \[\begin{aligned} (n-1)^{k+1} &= \sum_{m=0}^{k+1} \binom{k+1}{m} n^m (-1)^{k+1 -m} \\ &= n^{k+1} - (k+1)n^k + \sum_{m=0}^{k-1} \binom{k+1}{m} n^m (-1)^{k+1 -m} \end{aligned}\] So we have \[\begin{aligned} \lim_{n \to \infty} \frac{n^{k}}{n^{k+1} - (n-1)^{k+1}} & = \lim_{n \to \infty} \frac{n^{k}}{ (k+1)n^k - \sum_{m=0}^{k-1} \binom{k+1}{m} n^m (-1)^{k+1 -m} } \\ & = \lim_{n \to \infty} \frac{1}{ (k+1) - \sum_{m=0}^{k-1} \binom{k+1}{m} {\color{blue} n^{m-k} } (-1)^{k+1 -m} } \\ & = \frac{1}{ (k+1)}. \end{aligned}\] $$\tag*{$\blacksquare$}$$
Proposition. Let \((a_n)_{n \in \mathbb{N}}\) and \((b_n)_{n \in \mathbb{N}}\) be two sequences such that
\(b_n >0\), \(n \in \mathbb{N}\) and \(\lim\limits_{n \to \infty}b_n=+\infty\),
\(\lim\limits_{n \to \infty}\frac{a_n}{b_n}=g\).
then \[\lim_{n \to \infty}\frac{a_1+\ldots+a_n}{b_1+\ldots+b_n}=g.\]
Proof. We apply Stolz’s theorem with \(x_{n} = a_{1} + \ldots + a_{n}\) and \(y_{n} = b_{1} + \ldots + b_{n}\). Then the assumptions of the Stolz theorem are satisfied as \(y_{n+1} - y_{n} = b_{n+1} >0\) and \(y_{n} \ge b_{n}\) both diverge to \(+ \infty\), and \[\begin{aligned} \lim_{n \to \infty}\frac{x_n-x_{n-1}}{y_n-y_{n-1}} = \lim_{n \to \infty} \frac{a_n}{b_{n}} = g. \end{aligned}\] Therefore \(\lim\limits_{n \to \infty} \frac{x_n}{y_{n}} = g\) and the proof is finished. $$\tag*{$\blacksquare$}$$
Consider two sequences \((a_n)_{n \in \mathbb{N}}\) and \((b_n)_{n \in \mathbb{N}}\) defined by
\[\begin{aligned} {\color{red}a_n=\left(1+\frac{1}{n}\right)^n}, \qquad {\color{blue}b_n=\left(1+\frac{1}{n}\right)^{n+1}} \quad \text{ for all } \quad n \in \mathbb{N} \end{aligned}\].
We have the following properties.
Observe that \(a_n<b_n\) for all \(n \in \mathbb{N}\). Indeed, \[{\color{red}a_n=\left(1+\frac{1}{n}\right)^n}<{\color{blue}\left(1+\frac{1}{n}\right)^{n+1}=b_n,}\] since \(1<1+\frac{1}{n}\) for all \(n \in \mathbb{N}\).
The sequence \((a_n)_{n \in \mathbb{N}}\) is strictly increasing, i.e. \[a_n<a_{n+1} \quad \text{ for all } \quad n \in \mathbb{N}.\]
Proof. By the geometric-arithmetic mean inequality \(G_{n+1}<A_{n+1}\) (which is strict unless \(x_1=x_2=\ldots=x_{n+1}\)) with \[x_1=1\quad \text{ and } \quad x_2=x_3=\ldots=x_{n+1}=1+\frac{1}{n},\]
we obtain \[G_{n+1}=\left(\left(1+\frac{1}{n}\right)^{n}\right)^{1 / (n+1)} < \frac{1+n\left(1+\frac{1}{n}\right)}{n+1}=1+\frac{1}{n+1}=A_{n+1}.\]
Thus \[{\color{red}a_n}=\left(1+\frac{1}{n}\right)^n<\left(1+\frac{1}{n+1}\right)^{n+1}={\color{red}a_{n+1}}.\] $$\tag*{$\blacksquare$}$$
The sequence \((b_n)_{n \in \mathbb{N}}\) is strictly decreasing, i.e. \[b_{n+1}<b_n \quad \text{ for all } \quad n \in \mathbb{N}.\] Proof. By the harmonic-geometric mean inequality \(H_{n+1}<G_{n+1}\) (which is strict unless \(x_1=x_2=\ldots=x_{n+1}\)) with \[x_1=1\quad \text{ and }\quad x_2=x_3=\ldots=x_{n+1}=1+\frac{1}{n-1}=\frac{n}{n-1}.\] Then \[H_{n+1}=\frac{n+1}{1+n\frac{n-1}{n}}<\left(1+\frac{1}{n-1}\right)^{n / (n+1)}=G_{n+1},\] thus \[{\color{blue}b_n}=\left(1+\frac{1}{n}\right)^{n+1}<\left(1+\frac{1}{n-1}\right)^{n}={\color{blue}b_{n-1}}.\tag*{$\blacksquare$}\]
Collecting (1),(2),(3) we have \[2=a_1<a_n<a_{n+1}<b_{n+1}<b_n<b_1=4 \quad \text{ for all } \quad n \geq 2.\]
Thus the limits \(\lim_{n \to \infty} a_n\) and \(\lim_{n \to \infty}b_n\) exist and \[\lim_{n \to \infty}b_n=\lim_{n \to \infty}\left(1+\frac{1}{n}\right)a_n= \left(\lim_{n \to \infty}\left(1+\frac{1}{n}\right)\right)\left(\lim_{n \to \infty}a_n\right)=\lim_{n \to \infty}a_n.\]
Euler number. The limit of the sequences \((a_n)_{n \in \mathbb{N}}\) and \((b_n)_{n \in \mathbb{N}}\) is called the Euler number \[\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n=\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^{n+1}=e \simeq 2,718\ldots .\]
Fact. If \(\lim_{n \to \infty}a_n=+\infty\) or \(\lim_{n \to \infty}a_n=-\infty\), then \[\lim_{n \to \infty}\left(1+\frac{1}{a_n}\right)^{a_n}=e.\]
Proof. Let \(\lim_{n \to \infty}a_n=+\infty\) and consider \(b_n=\lfloor a_n \rfloor\). Then \(b_n \leq a_n<b_n+1\), hence \[\left(1+\frac{1}{b_n+1}\right)^{b_n}<\left(1+\frac{1}{a_n}\right)^{a_n}<\left(1+\frac{1}{b_n}\right)^{b_n+1}.\]
By the squeeze theorem it suffices to prove that \[\lim_{n \to \infty}\left(1+\frac{1}{b_n+1}\right)^{b_n}=\lim_{n \to \infty}\left(1+\frac{1}{b_n}\right)^{b_n+1}=e\] or even \[{\color{red}\lim_{n \to \infty}\left(1+\frac{1}{b_n}\right)^{b_n}=e}.\]
If \((b_n)_{n \in \mathbb{N}}\) were increasing then as a subsequence of \((n)_{n \in \mathbb{N}}\) we could conclude \(\lim_{n \to \infty}\left(1+\frac{1}{b_n}\right)^{b_n}=e\), since \(\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^{n}=e\).
But we only know that \(\lim_{n \to \infty}b_n=+\infty\). It does not mean that \((b_n)_{n \in \mathbb{N}}\) is increasing.
Let \(\varepsilon>0\) be given. Since \(\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^{n}=e\) we can find \(\widetilde{N}_{\varepsilon} \in \mathbb{N}\) so that \(n \geq \widetilde{N}_{\varepsilon}\) implies \[\left|\left(1+\frac{1}{n}\right)^n-e\right|<\varepsilon.\]
But \(\lim_{n \to \infty}b_n=+\infty\) thus we can find \(N_{\varepsilon} \in \mathbb{N}\) so that \(n \geq N_{\varepsilon}\) implies \(b_n \geq \widetilde{N}_{\varepsilon}\). In particular, we conclude that \[\left|\left(1+\frac{1}{b_n}\right)^{b_n}-e\right|<\varepsilon\] for all \(n \geq {N}_{\varepsilon}\) and thus
\[\lim_{n \to \infty}\left(1+\frac{1}{b_n}\right)^{b_n}=e.\]
Consequently, \(\lim_{n \to \infty}\left(1+\frac{1}{a_n}\right)^{a_n}=e\) as \(\lim_{n \to \infty}a_n=+\infty\).
Moreover,
\[\lim_{n \to \infty}\left(1-\frac{1}{a_n}\right)^{a_n}=e^{-1},\]
because
\[\lim_{n \to \infty}\left(1-\frac{1}{a_n}\right)^{a_n}=\lim_{n \to \infty}\frac{1}{\left(1+\frac{1}{a_n-1}\right)^{a_n}}=\frac{1}{e}.\] this implies \[\lim_{n \to \infty}\left(1+\frac{1}{a_n}\right)^{a_n}=e \quad \text{ if }\quad \lim_{n \to \infty}a_n=-\infty.\]
Exercise. Find \(\lim_{n \to \infty}\left(1+\frac{1}{2n}\right)^{4n}\).
Solution. Since \((2n)_{n \in \mathbb{N}}\) is a subsequence of \((n)_{n \in \mathbb{N}}\) we have \[\lim_{n \to \infty}\left(1+\frac{1}{2n}\right)^{2n}=e.\] Therefore, \[\begin{aligned} \lim_{n \to \infty}\left(1+\frac{1}{2n}\right)^{4n}&=\left(\lim_{n \to \infty}\left(1+\frac{1}{2n}\right)^{2n}\right)\left(\lim_{n \to \infty}\left(1+\frac{1}{2n}\right)^{2n}\right)\\ &=e \cdot e=e^2. \end{aligned}\] $$\tag*{$\blacksquare$}$$
Exercise. Find \(\lim_{n \to \infty}\left(1+\frac{1}{n^2+1}\right)^{4n^2+1}\).
Solution. Since \((n^2+1)_{n \in \mathbb{N}}\) is a subsequence of \((n)_{n \in \mathbb{N}}\) we have \[\lim_{n \to \infty}\left(1+\frac{1}{n^2+1}\right)^{n^2+1}=e.\] Therefore, \[\begin{aligned} &\lim_{n \to \infty}\left(1+\frac{1}{n^2+1}\right)^{4n^2+1}\\&=\left(\lim_{n \to \infty}\left(1+\frac{1}{n^2+1}\right)^{n^2+1}\right)^4\left(\lim_{n \to \infty}\left(1+\frac{1}{n^2+1}\right)^{-3}\right)=e^4.\end{aligned}\qquad\blacksquare\]
We write that \(s_n \ _{\overrightarrow{n \to \infty}}+\infty\) if for every \(M>0\) there is \(n \in \mathbb{N}\) such that \(n \geq N\) implies \(s_n \geq M\).
Similarly, \(s_n \ _{\overrightarrow{n \to \infty}}-\infty\) if for every \(M>0\) there is an integer \(N \in \mathbb{N}\) such that \(n \geq N\) implies \(s_n \leq -M\).
Upper limit and lower limit. Let \((s_n)_{n \in \mathbb{N}}\) be a sequence of real numbers.
The upper limit is defined by \[\limsup_{n \to \infty}s_n=\inf_{k \geq 1}\sup_{n \geq k}s_n.\]
The lower limit is defined by \[\liminf_{n \to \infty}s_n=\sup_{k \geq 1}\inf_{n \geq k}s_n.\]
Proposition. For a sequence \((s_n)_{n \in \mathbb{N}}\subset\mathbb R\), the upper and lower limits always exist.
Proof. Let \(\alpha_k=\sup_{n \geq k}s_n\). Then \({\color{blue}\alpha_{k+1} \leq \alpha_k}\) and \[\limsup_{n \to \infty}s_{n}=\inf_{k \geq 1}\sup_{n \geq k}s_n \underbrace{=}_{{\color{blue}(MCT)}}\lim_{n \to \infty}\alpha_k \text{ {\color{red} (possible infinite!)}}.\]
If \(\beta_k=\inf_{n \geq k}s_n\), then \({\color{blue}\beta_k \leq \beta_{k+1}}\) and \[\liminf_{n \to \infty}s_{n}=\sup_{k \geq 1}\inf_{n \geq k}s_n \underbrace{=}_{{\color{blue}(MCT)}}\lim_{n \to \infty}\beta_k \text{ {\color{red} (possible infinite!)}}\tag*{$\blacksquare$}.\]
Remark. We always have \(\beta_k=\inf_{n \geq k}s_n \leq \sup_{n \geq k}s_{n}=\alpha_k\). Thus
\[{\color{red}\liminf_{n \to \infty}a_n=\lim_{k \to \infty }\beta_k \leq \lim_{k \to \infty}\alpha_k = \limsup_{n \to \infty}s_n}.\]
Example 1. Consider \(a_n=(-1)^n\frac{n+1}{n}\). Let \[\beta_n=\sup\left\{(-1)^n\frac{n+1}{n},(-1)^{n+1}\frac{n+2}{n+1},\ldots\right\},\] then \[\beta_n=\begin{cases}\frac{n+1}{n} \text{ if }n \text{ is even,}\\ \frac{n+2}{n+1} \text{ if }n \text{ is odd.} \end{cases}\]
Thus \(\lim_{n \to \infty}\beta_n=1\). Therefore \[\limsup_{n \to \infty}a_n=1.\] Similarly \[\liminf_{n \to \infty}a_n=-1.\]
Example 2. Let \[a_n=\begin{cases}0 \text{ if }n \text{ is odd},\\ 1 \text{ if }n \text{ is even.} \end{cases}\] Then \[\begin{aligned} \beta_n&=\sup\left\{a_m\;:\;m \geq n\right\}=1, \\ \alpha_n&=\inf\{a_m\;:\;m \geq n\}=0. \end{aligned}\]
Therefore \[\begin{aligned} \limsup_{n \to \infty}a_n&=1,\\ \liminf_{n \to \infty}a_n&=0. \end{aligned}\]
Example 3. Let \(a_n=\frac{1}{n}\). Then \[\beta_n=\sup\left\{\frac{1}{m}\;:\;m \geq n\right\}=\frac{1}{n},\] so \(\lim_{n \to \infty}\beta_n=0.\) Similarly \[\alpha_n=\inf\left\{\frac{1}{m}\;:\;m \geq n\right\}=0,\] so \(\lim_{n \to \infty}\alpha_n=0\). Thus \[\limsup_{n \to \infty}a_n=\liminf_{n \to \infty}a_ n=0.\]
Definition. Let \((s_n)_{n \in \mathbb{N}}\) be a sequence of real numbers. We say that \(x \in \mathbb{R} \cup\{\pm \infty\}\) is an accumulation point of \((s_n)_{n \in \mathbb{N}}\) if \[s_{n_k} \ _{\overrightarrow{k \to \infty}}x\] for some subsequence \((s_{n_k})_{k \in \mathbb{N}}\).
Theorem. Let \((s_n)_{n \in \mathbb{N}}\) be a sequence of real numbers. Let \(E\) be the set of all accumulation points of \((s_n)_{n \in \mathbb{N}}\). Then \[\limsup_{n \to \infty}s_n=s^{*}=\sup E,\] \[\liminf_{n \to \infty}s_n=s_{*}=\inf E.\]
Suppose that \[\limsup_{n \to \infty}s_n=+\infty,\] thus \[\inf_{k \geq 1}\sup_{n \geq k}s_n=+\infty,\] so
\[\sup_{n \geq k}s_n=+\infty \quad \text{ for all }\quad k \in \mathbb{N}.\]
Hence there is \((n_k)_{k \in \mathbb{N}}\) so that
\[\lim_{k \to \infty}s_{n_k}=+\infty.\]
this gives \(s^{*}=\sup E=+\infty\). $$\tag*{$\blacksquare$}$$
Suppose that \[\limsup_{n \to \infty}s_n=-\infty,\]
so \[\lim_{k \to \infty}\sup_{n \geq k}s_n=-\infty.\]
This means that for every \(M>0\) there is \(N \in \mathbb{N}\) so that \(k \geq N\) implies \[\sup_{n \geq k}s_n \leq -M.\]
Hence \(s_{n} \leq -M\) for all \(n \geq N\), i.e. \[\lim_{n \to \infty}s_n=-\infty.\]
So \(E=\{-\infty\}\) and \(s^{*}=\sup E=-\infty\). $$\tag*{$\blacksquare$}$$
Claim:. Assume that \(\limsup_{n \to \infty}s_n=L\) and \(L \in \mathbb{R}\). Then
\(\sup E \leq L\),
\(L \in E\),
which implies \(s^{*}=\sup E=L\).
Remark:. This gives a stronger conclusion \[\limsup_{n \to \infty}s_n=\sup E=\max E.\]
Suppose that \(L <\sup E\). Thus there is \(x \in E\) such that \[L <x \leq \sup E,\] and there exists a sequence \((s_{n_j})_{j \in \mathbb{N}}\) so that \(\lim_{j \to \infty}s_{n_j}=x\), i.e. for every \(\varepsilon>0\) there exists \(K_0 \in \mathbb{N}\) so that \[j \geq K_0\quad \text{implies} \quad |s_{n_j}-x|<\varepsilon.\]
In particular, taking \({\color{red}\varepsilon=\frac{x-L}{2}}\) we obtain that \[{\color{blue}\frac{x+L}{2}=x-\varepsilon <s_{n_j} \quad \text{ for all } \quad j \geq K_0.}\]
Since \(\limsup_{n \to \infty}s_n=\inf_{k \geq 1}\sup_{n \geq k}s_n=L\) we obtain that for every \(\varepsilon>0\) there is \(K_{\varepsilon} \in \mathbb{N}\) so that \[k \geq K_{\varepsilon}\quad \text{implies}\quad L \leq \sup_{n \geq k}s_n<L+\varepsilon.\]
Taking \(\varepsilon=\frac{x-L}{2}\) we obtain \[\sup_{n \geq k}s_n<L+\varepsilon=\frac{x+L}{2}.\] Thus picking \(j_0 \geq K_0\) so that \(n_{j_0} \geq K_{\varepsilon}\) we obtain \[s_{n_{j_0}} \leq \sup_{n \geq K_{\varepsilon}}s_n<{\color{blue}\frac{L+x}{2}<s_{n_{j_0}}},\] which is impossible. Thus (a) must be true, i.e. \[\sup E \leq L.\]
We now construct \((s_{n_j})_{j \in \mathbb{N}}\) such that \(\lim_{j \to \infty}s_{n_j}=L\).
Since \(\limsup_{n \to \infty}s_n=\inf_{k \geq 1}\sup_{n \geq k}s_n=L\) then for any \(\varepsilon>0\) there is \(K_{\varepsilon} \in \mathbb{N}\) so that \(k \geq K_{\varepsilon}\) implies
(*). \[L \leq \sup_{n \geq k}s_n<L+\varepsilon.\]
Let \(\varepsilon=1\) and let \(K_1 \in \mathbb{N}\) so that (*) holds. Then there is \(n_1 \in \mathbb{N}\) such that \[L-1 \leq \sup_{n \geq K_1}s_n-1<s_{n_1}<\sup_{n \geq K_1}s_n<L+1.\]
Suppose that we have constructed inductively a sequence \(n_1<n_2<\ldots<n_j\) such that \[L-\frac{1}{j} \leq s_{n_j} \leq L+\frac{1}{j}.\]
We now construct \(n_{j+1}\). Set \(\varepsilon=\frac{1}{j+1}\) in (*) which yields a corresponding \(K_{1 / (j+1)} \in \mathbb{N}\). Let \(\widetilde{K}_j=\max(n_j,K_{1 / (j+1)})+1\). Using (*) we see \[L \leq \sup_{n \geq \widetilde{K}_j}s_n<L+\frac{1}{j+1}\] and we find \(n_{j+1}>\widetilde{K}_j>n_j\) such that \[L-\frac{1}{j+1} \leq \sup_{n \geq \widetilde{K}_j}s_n-\frac{1}{j+1}<s_{n_{j+1}} \leq \sup_{n \geq \widetilde{K}_j}s_n<L+\frac{1}{j+1}\] hence \[\lim_{j \to \infty}s_{n_j}=L.\] and we are done. $$\tag*{$\blacksquare$}$$
Proposition. A sequence \((s_n)_{n \in \mathbb{N}}\) is convergent and has a limit \(L \in \mathbb{R} \cup\{\pm \infty\}\) iff \[\liminf_{n \to \infty}s_n=\limsup_{n \to \infty}s_n.\]
Proof. If \(\lim_{n \to \infty}s_n=L\), then \(E=\{L\}\) thus \(s^{*}=s_{*}=L\) and by the previous theorem \[\liminf_{n \to \infty}s_n=\limsup_{n \to \infty}s_n=L.\]
Conversely, if \(\liminf_{n \to \infty}s_n=\limsup_{n \to \infty}s_n=L,\) then \[\alpha_k=\inf_{n \geq k}s_n \leq s_k \leq \sup_{n \geq k}s_n=\beta_k\]
and \(\lim_{k \to \infty}\alpha_k=\lim_{k \to \infty}\beta_k=L\), thus \(\lim_{n \to \infty}s_n=L\). $$\tag*{$\blacksquare$}$$