Convergence of a sequence. A sequence \((x_n)_{n \in \mathbb{N}}\) converges to a real number \(x \in \mathbb{R}\) if, for all \(\varepsilon>0\) there exists \(N_{\varepsilon} \in \mathbb{N}\) such that whenever \(n \geq N_{\varepsilon}\) it follows that \[|x-x_n|<\varepsilon.\]
Order limit theorem. Assume that \(\lim_{n \to \infty}a_n=a\) and \(\lim_{n \to \infty}b_n=b\).
If \(a_n \geq 0\) for all \(n \in \mathbb{N}\), then \(a \geq 0\).
If \(a_n \leq b_n\) for all \(n \in \mathbb{N}\), then \(a \leq b\).
If there is \(c \in \mathbb{R}\) so that \(c \leq b_n\) for each \(n \in \mathbb{N}\), then \(c \leq b\). Similarly, if \(a_n \leq c\) for all \(n \in \mathbb{N}\), then \(a \leq c\).
Squeeze Theorem. If \(x_n \leq y_n \leq z_n\) for all \(n \in \mathbb{N}\) and if \(\lim_{n \to \infty}x_n=\lim_{n \to \infty}z_n=L\), then \(\lim_{n \to \infty}y_n=L\).
Proof. Let \(\varepsilon>0\) be arbitrary, but fixed.
(*). \(\lim_{n \to \infty}x_n=L \iff\) for every \(\varepsilon_1>0\) there exists \(N_{\varepsilon_1}^1 \in \mathbb{N}\) so that \(n \geq N_{\varepsilon_1}^1\) implies \(|x_n-L|<\varepsilon_1\).
(*). \(\lim_{n \to \infty}z_n=L \iff\) for every \(\varepsilon_2>0\) there exists \(N_{\varepsilon_2}^2 \in \mathbb{N}\) so that \(n \geq N_{\varepsilon_2}^2\) implies \(|z_n-L|<\varepsilon_2\).
We apply (*) and (*) with \({\color{red}\varepsilon_1}={\color{blue}\varepsilon_2}=\varepsilon\), then for \(n \geq N_{\varepsilon}=\max(N_{\varepsilon_1}^1,N_{\varepsilon_2}^2)\) one has
\[{\color{red}(*) \iff L-\varepsilon <x_n<L+\varepsilon},\]
\[{\color{blue}(*) \iff L-\varepsilon <z_n<L+\varepsilon}.\] Since \(x_n \leq y_n \leq z_n\) for all \(n \in \mathbb{N}\) we obtain for \(n \geq N_{\varepsilon}\) that \[{\color{red}L-\varepsilon<x_n} \leq y_n \leq {\color{blue}z_n<L+\varepsilon}.\]
Thus if \(n \geq N_{\varepsilon}\), then \[|y_n-L|<\varepsilon,\] which proves that \(\lim_{n \to \infty}y_n=L\). $$\tag*{$\blacksquare$}$$
Exercise. Prove that \(\lim_{n \to \infty}\sqrt{n^2+1}-n=0\).
Solution. We will use the squeeze theorem. On the one hand, \[0 \leq \sqrt{n^2+1}-n.\] On the other hand, \[\sqrt{n^2+1}-n=\frac{(\sqrt{n^2+1}-n)(\sqrt{n^2+1}+n)}{\sqrt{n^2+1}+n}=\frac{n^2+1-n^2}{\sqrt{n^2+1}+n} \leq \frac{1}{n}.\] Since \(\lim_{n \to \infty}\frac{1}{n}=\lim_{n \to \infty}0=0\), by the squeeze theorem \[\lim_{n \to \infty}\sqrt{n^2+1}-n=0.\]$$\tag*{$\blacksquare$}$$
Theorem. For every \(a \in \mathbb{R}\) there is a sequence of rational numbers \((r_n)_{n \in \mathbb{N}}\) such that \[\lim_{n \to \infty}r_n=\alpha.\]
Proof. Recall the Dirichlet principle.
Theorem (Dirichlet). Let \(\alpha,Q\) be real numbers, \(Q \geq 1\). There exist \(a,q \in \mathbb{Z}\) such that \(1 \leq q \leq Q\) and \((a,q)=1\) and \[\left|\alpha-\frac{a}{q}\right| < \frac{1}{qQ} \leq \frac{1}{q^2}.\]
Applying Dirichlet’s theorem with \(Q=n \in \mathbb{N}\) one finds for each \(n \in \mathbb{N}\) integers \(a_n,q_n \in \mathbb{Z}\) such that \(1 \leq q_n \leq n\) and \((a_n, q_n)=1\) and \[\left|\alpha-\frac{a_n}{q_n}\right|<\frac{1}{q_nn}<\frac{1}{n}.\]
Now let \(\varepsilon>0\) be arbitrary but fixed and let \(N_{\varepsilon} \in \mathbb{N}\) be such that \(\frac{1}{N_{\varepsilon}}<\varepsilon\), then for every \(n \geq N_{\varepsilon}\) one has \[\frac{1}{n}\leq \frac{1}{N_{\varepsilon}}<\varepsilon.\]
thus taking \(r_n=\frac{a_n}{q_n} \in \mathbb{Q}\) and \(n \geq N_{\varepsilon}\) one gets \[|\alpha-r_n|<\frac{1}{n}<\varepsilon.\]
Hence \(\lim_{n \to \infty}r_n=\alpha\).$$\tag*{$\blacksquare$}$$
Increasing and decreasing sequences. A sequence of real numbers \((a_n)_{n \in \mathbb{N}}\) is
increasing if \(a_n \leq a_{n+1}\) for all \(n \in \mathbb{N}\);
decreasing if \(a_n \geq a_{n+1}\) for all \(n \in \mathbb{N}\).
Monotone sequence. A sequence is monotone if it is either increasing or decreasing.
Example.
\(\left(3+\frac{1}{n}\right)_{n \in \mathbb{N}}\) is decreasing, so it is monotone.
\(\left(n^{3}\right)_{n \in \mathbb{N}}\) is increasing, so it is monotone.
\(\left((-1)^n\right)_{n \in \mathbb{N}}\) is neither increasing nor decreasing, so it is not monotone.
Monotone convergence theorem (MCT). If a sequence is monotone and bounded then it converges.
Proof. Assume that \((x_n)_{n \in \mathbb{N}}\) is increasing and bounded. Consider the set \[E=\{x_n\colon n \in \mathbb{N}\} \subseteq \mathbb{R},\] which is nonempty and bounded. Let \({\color{blue}x=\sup E \in \mathbb{R}}\), which exists by the axiom of completeness (AoC). We will show that \(\lim_{n \to \infty}x_n={\color{blue}x}.\)
Let \(\varepsilon>0\) and note that there exists \(N_{\varepsilon} \in \mathbb{N}\) so that \[x-\varepsilon<x_{N_{\varepsilon}} \leq x.\]
But \((x_n)_{n \in \mathbb{N}}\) is increasing thus for any \(n \geq N_{\varepsilon}\) one has \[x-\varepsilon<x_{N_{\varepsilon}} \leq x_{n} \leq x<x+\varepsilon.\]
Hence \(|x_n-x|<\varepsilon\) for all \(n \geq N_{\varepsilon}\), which shows that \(\lim_{n \to \infty}x_n=x\). $$\tag*{$\blacksquare$}$$
Binomial theorem. For every \(n \in \mathbb{N}\) and \(x,y \in \mathbb{R}\) one has \[(x+y)^n=\sum_{k=0}^n {n \choose k}x^ky^{n-k}, \quad \text{ where } \quad {\color{red}{n \choose k}=\frac{n!}{k!(n-k)!},}\] and \(n!=1 \cdot 2 \cdot 3 \cdot \ldots \cdot n\), for all \(n \in \mathbb{N}\) and \(0!=1\).
Example for \(n=3\) :. \[(x+y)^3=x^3+3x^2y+3xy^2+y^3.\]
Example for \(n=5\) :. \[(x+y)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5.\]
Theorem.
If \(p>0\), then \(\lim_{n \to \infty}\frac{1}{n^p}=0\).
If \(p>0\), then \(\lim_{n \to \infty}\sqrt[n]{p}=1\)
\(\lim_{n \to \infty}\sqrt[n]{n}=1\).
If \(p>0\) and \(\alpha \in \mathbb{R}\), then \(\lim_{n \to \infty}\frac{n^{\alpha}}{(1+p)^n}=0\).
If \(|x|<1\), then \(\lim_{n \to \infty}x^n=0\).
Proof of (a): Take \(\varepsilon>0\) be arbitrary, but fixed. Then \[n>\left(\frac{1}{\varepsilon}\right)^{1 / p},\] which is possible by the Archimedian property.
Proof of (b): If \(p>1\) set \(x_n=\sqrt[n]{p}-1\), then \(x_n>0\) and by Bernoulli’s inequality \[1+nx_n \leq (1+x_n)^{n}=p,\]
so that
\[0<x_n \leq \frac{p-1}{n}.\]
But \[\lim_{n \to \infty}\frac{p-1}{n}=0,\]
thus by the squeeze theorem we conclude \[\lim_{n \to \infty}x_n=0\] as desired.
Proof of (c): Set \(x_n=\sqrt[n]{n}-1\). Then \(x_n \geq 0\) and by the binomial theorem \[n=(1+x_n)^n \geq {n \choose 2}x_n^2=\frac{n(n-1)}{2}x_n^2.\]
Hence \[0 \leq x_n \leq \left(\frac{2}{n-1}\right)^{1 / 2} \quad \text{ for } \quad n \geq 2.\]
But \[\lim_{n \to \infty}\left(\frac{2}{n-1}\right)^{1 / 2}=0.\] Thus by the squeeze theorem \[\lim_{n \to \infty}x_n=0\] as desired. $$\tag*{$\blacksquare$}$$
Proof of (d): Let \(k \in \mathbb{N}\) so that \(k>\alpha\). For \(n>2k\) by the binomial theorem \[(1+p)^n>{n \choose k}p^k=\frac{n(n-1)\ldots (n-k+1)}{k!}p^k> \frac{n^k p^k}{2^k k!},\]
since \(n \geq \frac{n}{2}, n-1 \geq \frac{n}{2},\ldots, n-k+1 \geq \frac{n}{2}\). Hence \[0<\frac{n^{\alpha}}{(1+p)^n}<\frac{2^kk!}{p^k}n^{\alpha-k} \quad \text{ for } \quad n > 2k.\]
Since \(\alpha-k<0\) thus \(\lim_{n \to \infty}n^{\alpha-k}=0\) by (a) and by the squeeze theorem \(\lim_{n \to \infty}\frac{n^{\alpha}}{(1+p)^{n}}=0\). $$\tag*{$\blacksquare$}$$
Proof of (e): Take \(\alpha=0\) in (d) and observe that if \(0<x<1\) then the sequence \(x_n=x^n\) is decreasing and bounded. Thus \(\lim_{n \to \infty}x_n=0\). $$\tag*{$\blacksquare$}$$
Proposition. If \(a>0\) and \(\lim_{n \to \infty}x_n=x_0\), then \(\lim_{n \to \infty}a^{x_n}=a^{x_0}\).
Proof. It suffices to prove that \(\lim_{n \to \infty}a^{x_n}=1\) if \(\lim_{n \to \infty}x_n=0\).
Assume \(a>1\). By the previous theorem we know that \[\lim_{n \to \infty}a^{1 / n}=\lim_{n \to \infty}a^{-1 / n}=1.\] Thus for any \(\varepsilon>0\) there is \(M_{\varepsilon} \in \mathbb{N}\) such that for any \(m \geq M_{\varepsilon}\) \[1-\varepsilon < a^{-1 / m}<a^{1 / m}<1+\varepsilon.\]
Now since \(\lim_{n \to \infty}x_n=0\) we find \(N_{m,\varepsilon}\in \mathbb{N}\) so that for \(n \geq N_{\varepsilon,m}\) \[|x_n|<\frac{1}{m} \iff -\frac{1}{m}<x_n<\frac{1}{m}.\]
Thus \[1-\varepsilon<a^{-1 / m}<a^{x_n}<a^{1 / m}<1+\varepsilon\] which proves \(|a^{x_n}-1|<\varepsilon\) for any \(n \geq N_{m,\varepsilon}\) proving that \[\lim_{n \to \infty}a^{x_n}=1.\]
If \(0<a<1\) we note that \[\lim_{n \to \infty}{a^{x_n}}=\lim_{n \to \infty}\frac{1}{\left(\frac{1}{a}\right)^{x_n}}\] and this completes the proof of the proposition. $$\tag*{$\blacksquare$}$$
Theorem. For all positive real numbers \(a_1,a_2,\ldots,a_n\) and all positive weights \(q_1,q_2,\ldots,q_n\) satisfying the following convexity condition \[{\color{red}q_1+\ldots+q_n=1,}\]
we have \[{\color{blue}a_1^{q_1}\cdot\ldots\cdot a_n^{q_n} \leq q_1a_1+\ldots+q_na_n.}\]
If \(q_1=q_2=\ldots=q_n=\frac{1}{n}\), then we have \[a_1^{q_1}\cdot\ldots\cdot a_n^{q_n}=(a_1\cdot\ldots\cdot a_n)^{1/n}\le \frac{a_1+\ldots+ a_n}{n}=q_1a_1+\ldots+q_na_n,\] which recovers the inequality between geometric and arithmetic means.
Proof: We first assume \[q_1,\ldots,q_n \in \mathbb{Q} \quad \text{ and } \quad q_1,\ldots,q_n>0.\]
We can assume that \({\color{red}q_i=\frac{k_i}{m}}\) for \(1 \leq i \leq n\) and \[{\color{red}k_1+\ldots+k_n=m.}\]
Invoking the inequality between geometric and arithmetic means we obtain \[\begin{aligned} \sum_{i=1}^{n}q_ia_i&=k_1\frac{a_1}{m}+\ldots+k_n\frac{a_n}{m} \\ &\geq m \left(\left(\frac{a_1}{m}\right)^{k_1}\cdot \ldots \cdot \left(\frac{a_n}{m}\right)^{k_n}\right)^{1 / m}\\ & = a_1^{k_1 / m}\cdot \ldots \cdot a_n^{k_n / m}\\ &=a_1^{q_1}\cdot \ldots \cdot a_n^{q_n}. \end{aligned}\]
If now all weights \(q_1,\ldots,q_n>0\) are real numbers, then for any \(1 \leq i \leq n\), we choose a sequence of positive rationals \((q_{i,k})_{k \in \mathbb{N}}\) so that \[\lim_{k \to \infty}q_{i,k}=q_i\] and so that \[{\color{red}\sum_{i=1}^{n}q_{i,k}=1} \quad \text{ for all }\quad k \in \mathbb{N}.\]
Then by the previous part \[a_1^{q_{i,k}}\cdot \ldots \cdot a_n^{q_{n,k}} \leq q_{1,k}a_1+\ldots+q_{n,k}a_{n}.\]
Passing with \(k \to \infty\) we conclude that \[{\color{blue}a_1^{q_1}\cdot\ldots\cdot a_n^{q_n} \leq q_1a_1+\ldots+q_na_n.}\tag*{$\blacksquare$}\]
Hölder’s inequality. Let \(1<p,q<\infty\) be such that \(\frac{1}{p}+\frac{1}{q}=1\). Then for any real numbers \(x_1,\ldots,x_n\) and \(y_1,\ldots,y_n\) one has \[\begin{aligned} \sum_{j=1}^n|x_jy_j| \leq \Big(\sum_{j=1}^n|x_j|^p\Big)^{1 / p}\Big(\sum_{j=1}^n|y_j|^q\Big)^{1 / q}. \end{aligned}\]
Proof. By the previous theorem for any \(a_1,b_1>0\) we have \[a_1^{\frac 1 p}b_1^{\frac 1 q} \leq \frac{1}{p}a_1+\frac{1}{q}b_1,\] which for \(a_1=a^p\) and \(b_1=b^q\) yields
(*). \[ab \leq \frac{1}{p}a^p+\frac{1}{q}b^q.\]
Let
\[a_j:=\frac{|x_j|}{\big(\sum_{j=1}^{n}|x_j|^{p}\big)^{1 / p}}, \ \ \ b_j:=\frac{|y_j|}{\big(\sum_{j=1}^{n}|y_j|^{q}\big)^{1 / q}}\].
Applying (*) to each \(1 \leq j \leq n\) one gets \[\begin{aligned} \sum_{j=1}^{n}a_jb_j &\leq \sum_{j=1}^{n}\left(\frac{1}{p}a_j^p+\frac{1}{q}b_j^{q}\right)\\& =\sum_{j=1}^{n}\left(\frac{|x_j|^p}{p\big(\sum_{j=1}^{n}|x_j|^{p}\big)}+\frac{|y_j^q|}{q\big(\sum_{j=1}^{n}|y_j|^{q}\big)}\right)\\&=\frac{1}{p}\frac{\sum_{j=1}^{n}|x_j|^p}{\sum_{j=1}^{n}|x_j|^p}+\frac{1}{q}\frac{\sum_{j=1}^{n}|y_j|^q}{\sum_{j=1}^{n}|y_j|^q}=\frac{1}{p}+\frac{1}{q}= 1. \end{aligned}\]
Thus we have proved \[\sum_{j=1}^{n}a_jb_j \leq 1,\]
which is equivalent to
\[\sum_{j=1}^n|x_jy_j| \leq \Big(\sum_{j=1}^n|x_j|^p\Big)^{1 / p}\Big(\sum_{j=1}^n|y_j|^q\Big)^{1 / q}\] and the proof of Hölder’s inequality is completed. $$\tag*{$\blacksquare$}$$