Cauchy sequences. A sequence \((a_n)_{n \in \mathbb{N}}\) is called a Cauchy sequence if for every \(\varepsilon>0\) there exists \(N_{\varepsilon} \in \mathbb{N}\) such that whenever \(m,n \geq N_{\varepsilon}\) it follows \[|a_n-a_m|<\varepsilon.\]
Convergent sequences. Recall that a sequence \((a_n)_{n \in \mathbb{N}}\) converges to \(a \in \mathbb{R}\) if for any \(\varepsilon>0\) there is \(N_{\varepsilon}\in \mathbb{N}\) such that whenever \(n \geq N_{\varepsilon}\) if follows \[|a_n-a|<\varepsilon.\]
Theorem. Every convergent sequence is a Cauchy sequence.
Proof. Let \(\varepsilon>0\) be given. If \[\lim_{n \to \infty}x_n=x,\]
then there is \(N_{\varepsilon} \in \mathbb{N}\) so that \(n \geq N_{\varepsilon}\) implies \[|x_n-x|<\frac{\varepsilon}{2}.\]
Thus for \(n,m \geq N_{\varepsilon}\) we obtain \[\begin{aligned} |x_m-x_n| \leq |x_n-x|+|x_m-x|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. \end{aligned}\] The proof is completed. $$\tag*{$\blacksquare$}$$
Lemma. Cauchy sequences are bounded.
Proof. Let \((x_n)_{n \in \mathbb{N}}\) be Cauchy. Given \(\varepsilon=1\) there is \(N \in \mathbb{N}\) so that if \(n,m \geq N\) then \(|x_n-x_m|<1\). Thus
\[|x_n| \leq |x_N|+1.\]
Taking \[M=\max\{|x_1|,|x_2|,\ldots,|x_N|,|x_N|+1 \}\]
we conclude \(|x_n| \leq M\) for all \(n \in \mathbb{N}\).$$\tag*{$\blacksquare$}$$
Bolzano–Weierstrass theorem. Every bounded sequence contains convergent subsequence.
Cauchy Criterion. A sequence \((x_n)_{n \in \mathbb{N}}\) converges iff it is a Cauchy sequence.
Proof: The implication \((\Longrightarrow)\) has already been proved. For the reverse implication \((\Longleftarrow)\) assume that \((x_n)_{n \in \mathbb{N}}\) is Cauchy. By the previous lemma the sequence is bounded. Hence by the Bolzano–Weierstrass theorem there is \((n_k)_{k \in \mathbb{N}}\) so that \[\lim_{k \to \infty}x_{n_k}=x \quad \text{ for some }\quad x\in \mathbb{R} \qquad {\color{red}(*)}.\] Let \(\varepsilon>0\) be given. Then there is \(N_{\varepsilon} \in \mathbb{N}\) so that \(n,m \geq N_{\varepsilon}\) implies \(|x_n-x_m|<\frac{\varepsilon}{2}\). By (*) we can choose \(n_k \in \mathbb{N}\) so that \(n_k \geq N_{\varepsilon}\) and \[|x_{n_k}-x|<\frac{\varepsilon}{2}.\] Then for \(n \geq N_{\varepsilon}\) and the triangle inequality \[|x_n-x| \leq |x_{n}-x_{n_k}|+|x_{n_k}-x| <\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. \qquad \tag*{$\blacksquare$}\]
Definition. We say that the series \(\sum_{n=1}^{\infty}a_n\) converges to \(A \in \mathbb{R}\) and write \(\sum_{n=1}^{\infty}a_n=A\) if the associated sequence of its partial sums \[s_n=\sum_{k=1}^{n}a_k \ _{\overrightarrow{n \to \infty}} \ A.\] If \((s_n)_{n \in \mathbb{N}}\) diverges the series \(\sum_{n=1}^{\infty}a_n\) is said to diverge.
Remark.
Saying that the series \(\sum_{n=1}^{\infty}a_n\) converges we understand that \(\left|\sum_{k=1}^{\infty}a_k\right|<\infty\).
Saying that the series \(\sum_{n=1}^{\infty}a_n\) diverges we understand that \(\left|\sum_{k=1}^{\infty}a_k\right|=\infty\).
Algebraic limit theorem for series. If \(\sum_{k=1}^{\infty}a_k=A\) and \(\sum_{k=1}^{\infty}b_k=B\) then \[\sum_{k=1}^{\infty}(\alpha a_k+\beta b_k)=\alpha A+\beta B.\]
Proof. Let \(A_n=\sum_{k=1}^na_k\) and \(B_n=\sum_{k=1}^nb_k\). We know that \[\lim_{n \to \infty}A_n=A, \quad \text{ and } \quad \lim_{n \to \infty}B_n=B,\]
so \[\begin{aligned} \lim_{n \to \infty}\sum_{k=1}^{n}(\alpha a_k+\beta b_k)&=\lim_{n \to \infty}\alpha \sum_{k=1}^{n}a_k+\beta\sum_{k=1}^{n}b_k \\ &=\alpha \lim_{n \to \infty}A_n+\beta \lim_{n \to \infty}B_n=\alpha A+\beta B.\qquad\end{aligned}\qquad\blacksquare\]
Geometric series. If \(0 \leq x<1\), then \(\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}.\) If \(x \geq 1\), the series diverges.
Solution. If \(x < 1\), then \[s_n=\sum_{k=0}^{n}x^k=\frac{1-x^{n+1}}{1-x}\] and the result follows if we let \(n \to \infty\).
For \(x \geq1\) note that \[\underbrace{1+1+\ldots+1}_{n} \leq s_n.\]
We have \(\lim_{n \to \infty}n=+\infty\), thus \(\lim_{n \to \infty}s_n=+\infty.\)
Theorem. The series \(\sum_{k=1}^{\infty}a_k\) converges iff for every \(\varepsilon>0\) there is \(N_{\varepsilon} \in \mathbb{N}\) such that whenever \(n>m \geq N_{\varepsilon}\) it follows \[\left|\sum_{k=m+1}^{n}a_k\right|<\varepsilon.\]
Proof. Let \(s_n=\sum_{k=1}^{n}a_k\) and we show that \((s_n)_{n\in\mathbb N}\) is a Cauchy sequence. Observe that whenever \(n>m \geq N_{\varepsilon}\) then \[|s_n-s_m|=\left|\sum_{k=m+1}^n a_k\right|<\varepsilon.\] We now apply the Cauchy Criterion for sequences and we are done.$$\tag*{$\blacksquare$}$$
Theorem. If the series \(\sum_{k=1}^{\infty}a_k\) converges then \(\lim_{n \to \infty}a_n=0\).
Proof. Let \(\varepsilon>0\) be given. Apply the previous theorem with \(m=n-1\), then \[|a_n|=|s_n-s_{n-1}|<\varepsilon\] whenever \(n>N_{\varepsilon}\), and we are done. $$\tag*{$\blacksquare$}$$
Remark. But \(\lim_{n \to \infty}a_n=0\) does not imply \(\left|\sum_{k=1}^{\infty}a_k\right|<\infty\).
Consider \(a_n=\frac{1}{n} \ _{\overrightarrow{n \to \infty}} \ 0\), but \(\sum_{n=1}^{\infty}\frac{1}{n}=\infty\).
Exercise. Determine if the series \[\sum_{n=1}^{\infty}(-1)^{n}\left(1-\frac{1}{n^3}\right)^{n^2}\] diverges or converges.
Solution. Since \((n^3)_{n \in \mathbb{N}}\) is a subsequence of \((n)_{n \in \mathbb{N}}\) we have \[\lim_{n \to \infty}\left(1-\frac{1}{n^3}\right)^{n^3}=e^{-1},\] hence \(\lim_{n \to \infty}\left(1-\frac{1}{n^3}\right)^{n^2}=1\), and the limit \(\lim_{n \to \infty}(-1)^{n}\left(1-\frac{1}{n^3}\right)^{n^2}\) does not exist, so the series diverges.$$\tag*{$\blacksquare$}$$
Comparison test. Assume that sequences \((a_k)_{k \in \mathbb{N}}\) and \((b_k)_{k \in \mathbb{N}}\) satisfy \[0 \leq a_k \leq b_k \quad \text{ for all }\quad k \in \mathbb{N}.\]
If \(\sum_{k=1}^{\infty}b_k\) converges, then \(\sum_{k=1}^{\infty}a_k\) converges.
If \(\sum_{k=1}^{\infty}a_k\) diverges, then \(\sum_{k=1}^{\infty}b_k\) diverges.
Proof. Both statements follows from the Cauchy Criterion for series: \[\left|\sum_{k=m+1}^{n}a_k\right|\leq \left|\sum_{k=m+1}^{n}b_k\right|.\] This completes the proof. $$\tag*{$\blacksquare$}$$
Exercise. Determine if the series \[\sum_{n=1}^{\infty}\frac{1}{n^2+\sqrt{n}+15}\] diverges or converges.
Solution. For all \(n \in \mathbb{N}\) we have \[\frac{1}{n^2+\sqrt{n}+15} \leq \frac{1}{n^2},\quad \text{ thus }\] \[\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty,\] hence \[\ \qquad \qquad \sum_{n=1}^{\infty}\frac{1}{n^2+\sqrt{n}+15}<\infty.\qquad \tag*{$\blacksquare$}\]
Exercise. Determine if the series \[\sum_{n=1}^{\infty}\frac{1}{\sqrt[3]{n}+\sqrt{n}+1}\] diverges or converges.
Solution. For all \(n \in \mathbb{N}\) we have \[\frac{1}{\sqrt[3]{n}+\sqrt{n}+1} \geq \frac{1}{3\sqrt{n}},\quad \text{ thus }\] \[\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}=\infty,\] hence \[\ \qquad \qquad \sum_{n=1}^{\infty}\frac{1}{\sqrt[3]{n}+\sqrt{n}+1}=\infty.\qquad \tag*{$\blacksquare$}\]
Theorem. A series of nonnegative terms \(a_k \geq 0\) converges iff its partial sums form a bounded sequence.
Proof. If \(\sum_{k=1}^{\infty}a_k<\infty\) one sees that \[s_N=\sum_{k=1}^{N}a_k \leq M=\sum_{k=1}^{\infty}a_k<\infty.\]
Conversely, we also know that \(s_N \leq s_{N+1} \leq M\) for all \(N \in \mathbb{N}\). Then the limit \[\lim_{N \to \infty}s_{N}\] exists by the (MCT).$$\tag*{$\blacksquare$}$$
We have seen that \[\sum_{n=1}^\infty\frac{1}{n}=\infty,\qquad \text{ and } \qquad \sum_{n=1}^\infty\frac{1}{n^2}<\infty.\]
Cauchy Condensation Test. Suppose that \((b_n)_{n \in \mathbb{N}}\) is decreasing and \(b_n \geq 0\) for all \(n \in \mathbb{N}\). Then the series \[{\color{red}\sum_{n=1}^{\infty}b_n<\infty} \quad \text{ converges }\] iff the series \[{\color{blue}\sum_{n=1}^{\infty}2^n b_{2^n}<\infty} \quad \text{ converges.}\]
Corollary. The series \[\sum_{n=1}^{\infty}\frac{1}{n^p} <\infty \quad \text{ iff }\quad p>1\]
Proof. The sequence \(b_n=\frac{1}{n^p}\) is decreasing and \(b_n \geq 0\) for all \(n \in \mathbb{N}\). By the Cauchy condensation test we obtain \[\sum_{n=1}^{\infty}\frac{1}{n^p}<\infty \quad \iff \quad \sum_{n=0}^{\infty}\frac{2^n}{2^{pn}}<\infty.\]
But the latter converges provided that
\[\sum_{n=0}^{\infty}\frac{2^n}{2^{pn}}=\sum_{n=0}^{\infty}2^{(1-p)n}=\frac{1}{1-\frac{1}{2^{p-1}}}<\infty \quad \iff \quad p>1.\tag*{$\blacksquare$}\]
Theorem. \[\sum_{n=0}^{\infty}\frac{1}{n!}=e.\]
Proof. Let \(s_n=\sum_{k=0}^{n}\frac{1}{k!}\). Then
\(s_n<s_{n+1}\) for all \(n \in \mathbb{N}\),
\(s_{n}=\sum_{k=0}^{n}\frac{1}{k!}=1+1+\sum_{k=2}^{n}\frac{1}{k!}<2+\sum_{k=2}\frac{1}{2^{k-1}}<3\).
Thus the limit \(\lim_{n \to \infty}s_n\) exists.
Let \(t_n=\left(1+\frac{1}{n}\right)^{n}\), then \(\lim_{n \to \infty}t_n=e\). By the binomial theorem
\[t_n=\left(1+\frac{1}{n}\right)^{n}=\sum_{k=0}^{n}{n \choose k}\frac{1}{n^k}.\].
Then \[\begin{aligned} t_n&=\left(1+\frac{1}{n}\right)^{n}=\sum_{k=0}^{n}{n \choose k}\frac{1}{n^k} \\&=\sum_{k=0}^{n}\frac{n(n-1)\cdots (n-k+1)}{k!}\frac{1}{n^k} \\&=1+1+\frac{1}{2!}\left(1-\frac{1}{n}\right)+\frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)+\ldots\\&+\frac{1}{n!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdot \ldots \cdot \left(1-\frac{n-1}{n}\right) \le \sum_{k=0}^{n}\frac{1}{k!}=s_n. \end{aligned}\] Thus
\[e=\lim_{n \to \infty}t_n \leq \lim_{n \to \infty}s_n.\]
Next if \(n \geq m\) \[t_n \geq 1+1+\frac{1}{2!}\left(1-\frac{1}{n}\right)+\ldots+\frac{1}{m!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{m-1}{n}\right).\]
Let \(n \to \infty\) keeping \(m\) fixed, we get \[e=\lim_{n \to \infty}t_n \geq \sum_{k=0}^{m}\frac{1}{k!}.\]
Letting \(m \to \infty\) we see \(\lim_{m \to \infty}s_m \leq e\). \[\lim_{m \to \infty}s_m=\lim_{m \to \infty}\sum_{k=0}^{m}\frac{1}{k!}=e.\] This completes the proof of the theorem. $$\tag*{$\blacksquare$}$$
We have \(s_n=\sum_{k=0}^{n}\frac{1}{k!}<e\) for all \(n \in \mathbb{N}\). Indeed \[\begin{aligned} e-s_n&=\sum_{k=n+1}^{\infty}\frac{1}{k!}=\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\ldots \\&=\frac{1}{(n+1)!}\left(1+\frac{1}{n+2}+\frac{1}{(n+2)(n+3)}+\ldots\right) \\&<\frac{1}{(n+1)!}\left(1+\frac{1}{n+1}+\frac{1}{(n+1)^2}+\ldots\right) \\&\leq \frac{1}{(n+1)!}\frac{1}{1-\frac{1}{n+1}}= \frac{1}{(n+1)!}\frac{n+1}{n}=\frac{1}{n! n}. \end{aligned}\] Hence we conclude
The error estimate (*). \[0<e-s_n<\frac{1}{n! n}.\]
Theorem. The Euler number \(e\) is irrational.
Proof. Suppose \(e\) is rational. Then \({\color{red}e=\frac{p}{q}}\) where \(p,q \in \mathbb{N}\). By (*) we have \[0<q!(e-s_q)<\frac{1}{q}.\] By our assumption \[{\color{red}q!e\in\mathbb N \quad \text{ is an integer}.}\]
Since \[q!s_{q}=q!\left(1+1+\frac{1}{2!}+\ldots+\frac{1}{q!}\right) \in \mathbb{N},\] we see \(q!(e-s_q) \in \mathbb{N}\), but if \(q > 1\) and this is impossible since \[0<q!(e-s_q)<1 / q<1.\] Hence \(e\) must be irrational. $$\tag*{$\blacksquare$}$$
We know that
\[\lim_{n \to \infty}\left(1+\frac{x}{n}\right)^n=e^x.\]. for any \(x \in \mathbb{R}\).
Also
\[\sum_{n=0}^{\infty}\frac{1}{n!}=e.\].
Theorem. Let \(x \in \mathbb{R}\), then \[\sum_{n=0}^{\infty}\frac{x^n}{n!}=e^x.\]
Proof. Let \(S_n=\sum_{k=0}^{n}\frac{x^k}{k!}\), then by the binomial theorem we may write
\[\begin{aligned} \left|S_n-\left(1+\frac{x}{n}\right)^n\right|&=\left|\sum_{k=2}^n \left(1-\left(1-\frac{1}{n}\right)\cdot \ldots \cdot \left(1-\frac{k-1}{n}\right)\right)\frac{x^k}{k!}\right|\\&\leq \sum_{k=2}^n \left(1-\left(1-\frac{1}{n}\right)\cdot \ldots \cdot \left(1-\frac{k-1}{n}\right)\right)\frac{|x|^k}{k!}. \end{aligned}\] Let us also note that \[\left(1-\frac{1}{n}\right)\cdot \ldots \cdot \left(1-\frac{k-1}{n}\right) \geq 1-\sum_{j=1}^{k-1}\frac{j}{n}=1-\frac{k(k-1)}{2n}\] for \(2 \leq k \leq n\).
Thus \[\left|S_n-\left(1+\frac{x}{n}\right)^n\right| \leq \sum_{k=2}^n\frac{k(k-1)}{2n}\frac{|x|^k}{k!}= \frac{1}{2n}\sum_{k=2}^{n}\frac{|x|^k}{(k-2)!}.\]
Using the Stolz theorem \[\begin{aligned} \lim_{n \to \infty}\frac{1}{2n}\sum_{k=2}^{n}\frac{|x|^k}{(k-2)!}&=\lim_{n \to \infty}\frac{\sum_{k=2}^{n+1}\frac{|x|^k}{(k-2)!}-\sum_{k=2}^{n}\frac{|x|^k}{(k-2)!}}{(2n+2)-2n}\\ &=\lim_{n \to \infty}\frac{1}{2}\frac{|x|^{n+1}}{(n-1)!}=0. \end{aligned}\]
Thus \[\lim_{n \to \infty}S_n=\lim_{n \to \infty}\left(1+\frac{x}{n}\right)^n=e^x\] as desired. $$\tag*{$\blacksquare$}$$