Divergence of a sequence. We say that a sequence \((a_n)_{n \in \mathbb{N}}\) diverges to \(+\infty\) and write \[\lim_{n \to \infty}a_n=+\infty\] iff for any \(M>0\) there exists \(N_{M} \in \mathbb{N}\) such that for all \(n \geq N_{M}\) \[a_n>M.\]
We have a similar definition for \(\lim_{n \to \infty}a_n=-\infty\).
Example. \((n^2)_{n \in \mathbb{N}}\) diverges to \(+\infty\), whereas \((\sqrt{n}-n)_{n \in \mathbb{N}}\) diverges to \(-\infty\).
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 increasing, 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 \[2=a_1<a_n<b_n<b_1=4 \quad \text{ for all } \quad n \geq 2.\]
By the (MCT) 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 \[\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n=\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^{n+1}=e \simeq 2,718\ldots .\]
Definition. Let \((a_n)_{n \in \mathbb{N}}\) be a sequence of real numbers, and \(n_1<n_2<\ldots<n_k<\ldots\) be an increasing sequence of positive integers. Then the sequence \[(a_{n_1},a_{n_2},\ldots,a_{n_k},\ldots)\] is called a subsequence of \((a_n)_{n \in \mathbb{N}}\) and is denoted by \((a_{n_k})_{k \in \mathbb{N}}\).
Example. Let \((a_n)_{n \in \mathbb{N}}=\left(1,\frac{1}{2},\frac{1}{3},\frac{1}{4}\ldots,\right)\), then \(\big(\frac{1}{2},\frac{1}{4},\frac{1}{6},\ldots\big) \text{ and }\big(\frac{1}{10},\frac{1}{100},\frac{1}{1000},\ldots\big)\) are subsequences of \((a_n)_{n \in \mathbb{N}}\). The sequences \[\left(\frac{1}{10},\frac{1}{2},\frac{1}{100},\ldots\right) \quad \text{ and } \quad (1,1,\ldots)\quad \text{ are {\color{red}NOT!}}.\]
Theorem. Subsequences of a convergent sequence converge to the same limit as the original sequence.
Proof. Assume \(\lim_{n \to \infty}a_n=a\) and let \((a_{n_k})_{k \in \mathbb{N}}\) be a subsequence. Given \(\varepsilon>0\) there is \(N_{\varepsilon} \in \mathbb{N}\) so that \[n \geq N_{\varepsilon}\qquad \text{ implies } \quad |a_n-a|<\varepsilon.\]
Because \(n_k \geq k\) for all \(k \in \mathbb{N}\), the same \(N_{\varepsilon}\) will suffice for the subsequence, that is \[|a_{n_k}-a|<\varepsilon \quad \text{ whenever }\quad k \geq N_{\varepsilon}.\] $$\tag*{$\blacksquare$}$$
Fact. If \(\lim_{n \to \infty}a_n=+\infty\) or \(\lim_{n \to \infty}a_n=-\infty\), then \[\lim_{n \to \infty}\left(1+\frac{1}{a_n}\right)^{a_n}=e.\]
In particular, \(\lim_{n \to \infty}\left(1+\frac{x}{n}\right)^{n}=e^x\) for any \(x \in \mathbb{N}\).
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, \(\lim_{n \to \infty}\left(1+\frac{1}{a_n}\right)^{a_n}\) 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 \[\lim_{n \to \infty}\left(1+\frac{1}{a_n}\right)^{a_n}=e \quad \text{ if }\quad \lim_{n \to \infty}a_n=-\infty.\]
For the second part we take \[{\color{red}a_n=\frac{n}{x}},\] then either \[\lim_{n \to \infty}a_n=+\infty \quad \text{ or }\quad \lim_{n \to \infty}a_n=-\infty.\]
Hence \[\lim_{n \to \infty}\left[\left(1+\frac{1}{a_n}\right)^{a_n}\right]^x=e^x.\]
Here we have used the following simple fact: if \(\lim_{n \to \infty}a_n =a\), then for any \(\alpha \in \mathbb{R}\) we have \[\lim_{n \to \infty}a_n^{\alpha}=a^{\alpha}.\] Prove it!
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, \[\begin{aligned} &\lim_{n \to \infty}\left(1+\frac{1}{n^2+1}\right)^{4n^2+1}\\&=\left(\lim_{n \to \infty}\left(1+\frac{1}{n^2+1}\right)^{n^2+1}\right)^4\left(\lim_{n \to \infty}\left(1+\frac{1}{n^2+1}\right)^{-3}\right)=e^4.\end{aligned}\qquad\blacksquare\]
Convergence of a series. Let \((b_n)_{n \in \mathbb{N}}\) be a sequence. An infinite series is a formal expression of the form \[\sum_{n=1}^{\infty}b_n=b_1+b_2+b_3+\ldots\]
We define the corresponding sequence of partial sums \((s_n)_{n \in \mathbb{N}}\) by \[{\color{blue}s_m=\sum_{n=1}^{m}}b_n=b_1+b_2+\ldots+b_m.\]
We say that \(\sum_{n=1}^{\infty}b_n\) converges to \(B\) if \[\lim_{n \to \infty}s_n=B.\]
Exercise. 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 \(\lim_{n \to \infty}s_n=+\infty.\)
Exercise. \[\sum_{n=1}^{\infty}\frac{1}{k^2}<\infty.\]
Solution. Because the terms in the sum are all positive the sequence \[s_n=\sum_{k=1}^{n}\frac{1}{k^2} \quad \text{ is increasing.}\]
We now show that \((s_n)_{n \in \mathbb{N}}\) is bounded.
Then the (MCT) will prove that the series converges.
To prove boundedness of \((s_n)_{n \in \mathbb{N}}\) we note that \[\begin{aligned} s_n&=1+\frac{1}{2 \cdot 2}+\frac{1}{3 \cdot 3}+\frac{1}{4 \cdot 4}+\ldots+\frac{1}{n \cdot n} \\&<1+\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{(n-1)n} \\ &= 1+\left(1-{\color{red}\frac{1}{2}}\right)+\left({\color{red}\frac{1}{2}}-{\color{blue}\frac{1}{3}}\right)+\left({\color{blue}\frac{1}{3}}-{\color{purple}\frac{1}{4}}\right)+\ldots+\left(\frac{1}{n-1}-\frac{1}{n}\right) \\&=2-\frac{1}{n}<2. \end{aligned}\]
Thus by the (MCT) the limit \(\lim_{n \to \infty}s_n\) exists.$$\tag*{$\blacksquare$}$$
One can also prove that \({\color{blue}\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}}\). This is also Euler’s result.
Harmonic series. \[\sum_{n=1}^{\infty}\frac{1}{n}=\infty.\]
Solution. Note that \[\begin{aligned} 1+\frac{1}{2}+&{\color{red}\left(\frac{1}{3}+\frac{1}{4}\right)}+{\color{blue}\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)}+{\color{purple}\left(\frac{1}{9}+\ldots+\frac{1}{16}\right)}+{\color{brown}\left(\frac{1}{17}+\ldots\right.} \\& \geq 1+\frac{1}{2}+{\color{red}2\cdot \frac{1}{4}}+{\color{blue}4 \cdot \frac{1}{8}}+{\color{purple}8\cdot \frac{1}{16}}+{\color{brown}16 \cdot \frac{1}{32}}+\ldots \\& =1+\frac{1}{2}+{\color{red}\frac{1}{2}}+{\color{blue}\frac{1}{2}}+{\color{purple}\frac{1}{2}}+{\color{brown}\frac{1}{2}}+\ldots =1+\lim_{n \to \infty}\frac{n}{2}=\infty. \end{aligned}\] Thus \(s_n=\sum_{k=1}^{n}\frac{1}{k} \ _{\overrightarrow{n \to \infty}}\ \infty\). $$\tag*{$\blacksquare$}$$
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 \[{\color{blue}\sum_{n=1}^{\infty}2^n b_{2^n}<\infty} \quad \text{ converges.}\]
Proof. Let \[s_n=b_1+b_2+\ldots+b_n,\] \[t_k=b_1+2b_2+\ldots+2^kb_{2^k}.\]
For \(n<2^k\) one has \[\begin{aligned} s_n &\leq b_1+\overbrace{b_2+b_3}^{2}+\ldots+\overbrace{b_{2^k}+\ldots+b_{2^{k+1}-1}}^{2^k}\\&\leq b_1+2b_2+\ldots+2^k b_{2^k}=t_k. \end{aligned}\]
(*). so that \(s_n \leq t_k\) for \(n<2^k\).
If \(n>2^k\) one has \[\begin{aligned} s_n &\geq b_1+b_2+(b_3+b_4)+\ldots+(b_{2^{k-1}+1}+\ldots+b_{2^k}) \\& \geq \frac{1}{2}b_1+b_2+2b_4+\ldots+2^{k-1}b_{2^k}=\frac{1}{2}t_k. \end{aligned}\]
(**). Thus \(2s_n \geq t_k\) for \(n>2^k\).
By (*) and (**) the sequences \((s_n)_{n \in \mathbb{N}}\) and \((t_k)_{k \in \mathbb{N}}\) are either both bounded or both unbounded. $$\tag*{$\blacksquare$}$$
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 \[\sum_{n=1}^{\infty}\frac{1}{n^p}<\infty \quad \iff \quad \sum_{n=1}^{\infty}\frac{2^n}{2^{pn}}<\infty.\]
But the latter converges provided that
\[\sum_{n=1}^{\infty}\frac{2^n}{2^{pn}}=\sum_{n=1}^{\infty}2^{(1-p)n}=\frac{1}{1-\frac{1}{2^{p-1}}}<\infty \quad \iff \quad p>1.\tag*{$\blacksquare$}\]
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 \[\begin{aligned} t_n&=\left(1+\frac{1}{n}\right)^{n}=\sum_{k=0}^{n}{n \choose k}\frac{1}{n^k} \\&=\sum_{k=0}^{n}\frac{n(n-1)\cdots (n-k+1)}{k!}\frac{1}{n^k} \\&=1+1+\frac{1}{2!}\left(1-\frac{1}{n}\right)+\frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)+\ldots\\&+\frac{1}{n!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdot \ldots \cdot \left(1-\frac{n-1}{n}\right) =\sum_{k=0}^{n}\frac{1}{k!}=s_n. \end{aligned}\] 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. $$\tag*{$\blacksquare$}$$