Sequences

Let $0<x<1$, then

\[x = 0,\bar{a_{1}}\bar{a_{2}}\bar{a_{3}}\dots\]

Let $x_{1} = 10x$, then $0<x_{1} <10$. If $0 \leq x_{1} <1$, then the first decimal digit is given by $a_{1} = 0$. If not, iterate to find the natural number $0 \leq n_{1} \leq 9$ such that $n_{1} \leq x_{1} < n_{1} + 1.$ Then $0 \leq a_{1}=n_{1} \leq 9 \quad \text{(1)}$.

Now take $x_{2} = 10(x_{1}-n_{1})$. Note that $0\leq x_{2} < 10$. Take $a_{2} = n_{2}$, where $n_{2}$ is such that $n_{2}\leq x_{2} \leq n_{2}+1.$ From (1), we have that

\[\begin{aligned} n_{1} &\leq 10x < n_{1} + 1\\ \implies \frac{n_{1}}{10} &\leq x < \frac{n_{1}}{10}+\frac{1}{10} \\ \implies 0 &\leq x-\frac{n_{1}}{10} < \frac{1}{10}. \end{aligned}\]

Moreover,

\[\begin{aligned} &n_{2} \leq 10(x_{1}-n_{1}) < n_{2} + 1\\ \implies & \frac{n_{2}}{10} \leq x_{1}-n_{1} < \frac{n_{2}}{10}+\frac{1}{10} \\ \implies & n_{1}+\frac{n_{2}}{10} \leq x_{1} < n_{1} + \frac{n_{2}}{10} + \frac{1}{10} \\ \implies & n_{1}+\frac{n_{2}}{10} \leq 10x < n_{1} + \frac{n_{2}}{10} + \frac{1}{10} \\ \implies & \frac{n_{1}}{10}+\frac{n_{2}}{100} \leq x < \frac{n_{1}}{10} + \frac{n_{2}}{100} + \frac{1}{100} \\ \implies& 0\leq x -\frac{n_{1}}{10} -\frac{n_{2}}{100} < \frac{1}{10^{2}}. \end{aligned}\]

Iterating the process, we obtain

\[0\leq n_{1},n_{2},n_{3},...,n_{k} \leq 9,\]

with

\[\frac{n_{1}}{10} +\frac{n_{2}}{10^{2}}+\frac{n_{3}}{10^{3}} + \dots + \frac{n_{k}}{10^{k}} \leq x \leq \frac{n_{1}}{10} +\frac{n_{2}}{10^{2}}+\frac{n_{3}}{10^{3}} + \dots + \frac{n_{k}}{10^{k}} + \frac{1}{10^{k}}.\]

Let $x_{k+1} = 10(x_{k}-n_{k})$. Then there exists $n_{k+1} \in \mathbb{N}$ such that $n_{k+1} \leq x_{k+1} < n_{k+1} +1,$ with $0\leq n_{k+1} \leq 9.$ Since

\[0 \leq \underbrace{ x-\frac{n_{1}}{10}-\frac{n_{2}}{10^{2}} - \frac{n_{3}}{10^{3}} - \dots - \frac{n_{k}}{10^{k}} }_{ y_{1} } < \frac{1}{10^{k}},\]

we have that $0 \leq 10^{k+1} y_{1} < 10$. Then $x_{k+1} = 10^{k+1}y_{1}$. We know that $n_{k+1} \leq x_{k+1} < n_{k+1} +1,$, so

\[\begin{aligned} & n_{k+1} \leq 10^{k+1}\left( x-\frac{n_{1}}{10}-\frac{n_{2}}{10^{2}}-\frac{n_{3}}{10^{3}}-\dots-\frac{n_{k}}{10^{k}} \right) < n_{k+1}+1\\ \implies & 0 \leq x-\frac{n_{1}}{10}-\frac{n_{2}}{10^{2}}-\frac{n_{3}}{10^{3}}-\dots-\frac{n_{k}}{10^{k}} - \frac{n_{k+1}}{10^{k+1}} < \frac{1}{10^{k+1}}. \end{aligned}\]

Conclude by induction that $x - \left( \frac{n_{1}}{10} + \frac{n_{2}}{10^{2}} +\frac{n_{3}}{10^{3}} +\dots+ \frac{n_{k}}{10^{k}} \right) < \frac{1}{10^{k}}$. Sequences are born in order to approximate numbers.

Definition (Real-valued sequence)

A sequence in $\mathbb{R}$ (denoted $\{x_{n}\}_{n=1}^\infty$) is a function $\underset{n \to x_{n}}{f:\mathbb{N} \to \mathbb{R}}$.

Definition (Convergence)

We say that a sequence $\{x_{n}\}_{n=1}^\infty$ converges to $L$ $(\lim_{ n \to \infty } x_{n} = L)$, denoted by $x_{n} \underset{n \rightarrow \infty}{\longrightarrow} L$, if for every $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that for every $n \geq N$, $\lvert x_{n}-L \rvert < \varepsilon$.

Example

Given $x \in (0,1)$, there exists $x_{m} = \sum_{i=1}^{m} \frac{n_{i}}{10^{i}}$ satisfying $\lvert x-x_{m} \rvert < \frac{1}{10^{m}}$. Let $\varepsilon > 0$ be fixed and arbitrary. Take $N = \left\lceil \log_{10}\left( \frac{1}{\varepsilon} \right) \right\rceil$. Let $n \geq N$ be fixed and arbitrary. Note that

\[\log_{10} \left( \frac{1}{\varepsilon} \right) \leq M \leq n\] \[\implies\quad\frac{1}{\varepsilon} \leq 10^{n}\] \[\implies \frac{1}{10^{n}} \leq \varepsilon\] \[\implies \lvert x_{n} - x \rvert < \frac{1}{10^{n}} \leq \varepsilon.\]

Conclude that $\lim_{ n \to \infty } x_{n} = x$.

Theorem (Uniqueness of the limit)

Let $\{x_{n}\}_{n=1}^\infty$ be a sequence such that $x_{n} \underset{n \rightarrow \infty}{\longrightarrow} \ell_{1}$ and $x_{n} \underset{n \rightarrow \infty}{\longrightarrow} \ell_{2}$. Then $\ell_{1} = \ell_{2}$.

Proof: Suppose for the sake of contradiction that $\ell_{1} \neq \ell_{2}$. Consider

\[0 < \lvert \ell_{2} - \ell_{2} \rvert \leq \lvert \ell_{1}-x_{n} \rvert + \lvert x_{n}-\ell_{2} \rvert.\]

Take $\varepsilon = \frac{\lvert \ell_{1}-\ell_{2} \rvert}{2}$. Then there exists $N \in \mathbb{N}$ such that $\lvert \ell_{1}-x_{n} \rvert < \varepsilon$ and $\lvert x_{n} - \ell_{2} \rvert < \varepsilon$ for $n \geq N$. Hence,

\[0 < \lvert \ell_{1} - \ell_{2} \rvert \leq \lvert x_{n}-\ell_{1} \rvert + \lvert x_{n} - \ell_{2} \rvert < 2\varepsilon = \lvert \ell_{1} - \ell_{2} \rvert. \quad (\Rightarrow\!\Leftarrow)\]

Theorem (Boundedness)

Let $\{x_{n}\}_{n=1}^\infty$ be such that $x_{n} \underset{n \rightarrow \infty}{\longrightarrow} L$. Then $\{x_{n}\}_{n=1}^\infty$ is bounded.

Proof: Take $\varepsilon = 1$. Then there exists $N \in \mathbb{N}$ such that $\lvert x_{n}-1 \rvert < 1$ for $n \geq N$, that is,

\[L -1 < x_{n} <L+1 \implies \lvert x_{n} \rvert < \max\{\lvert L+1 \rvert , \lvert L-1 \rvert \}.\]

Take $M = \max \{\lvert x_{1} \rvert\ , \lvert x_{2} \rvert,\dots,\lvert x_{N} \rvert, \lvert L-1 \rvert, \lvert L+1 \rvert\}$. Then $\lvert x_{n} \rvert \leq M$ for every $n \in \mathbb{N}$.

Corollary

Let $\{x_{n}\}_{n=1}^\infty$ be an unbounded sequence. Then $\{x_{n}\}_{n=1}^\infty$ is divergent.

Proof: This follows directly from the contrapositive of the theorem.

Example

Let $b>1$ and let $y_{n} = b^{n}$ for $n \geq 0$. Assume that there exists $M \in \mathbb{R}$ such that $b^{n} \leq M$ for every $n \in \mathbb{N}$. By Bernoulli’s inequality, we have that $1 + n(b-1) \leq b^{n} \leq M$. But this implies that $n \leq \frac{M-1}{b-1}$ for every $n \in \mathbb{N}$, that is, that the sequence $\{n_{}\}_{n=1}^\infty$ is bounded $(\Rightarrow\!\Leftarrow)$.

Definition (Cauchy sequence)

Let $\{x_{n}\}_{n=1}^\infty \subseteq \mathbb{R}$. We say that the sequence is Cauchy if for every $\varepsilon>0$ there exists $N_{0} \in \mathbb{N}$ such that $\lvert x_{n}-x_{m} \rvert < \varepsilon$ for all $n,m \geq N_{0}$.

Theorem (Convergence implies Cauchy)

Let $\{x_{n}\}_{n=1}^\infty$ be a sequence such that $x_{n} \underset{n \rightarrow \infty}{\longrightarrow} L \in \mathbb{R}$. Then $\{x_{n}\}_{n=1}^\infty$ is Cauchy.

Proof: Assume that $x_{n} \underset{n \rightarrow \infty}{\longrightarrow} L$. Then, given $\varepsilon > 0$, there exists $N_{0} \in \mathbb{N}$ such that $\lvert x_{n} - L \rvert > \frac{\varepsilon}{2}$ for every $n \geq N_{0}$. Let $n,m \geq N_{0}$ be fixed and arbitrary. Then

\[\lvert x_{n} - x_{m} \rvert \leq \lvert x_{n} - L \rvert + \lvert x_{m} - L \rvert < \varepsilon.\]

Conclude that $\{x_{n}\}_{n=1}^\infty$ is Cauchy.

Corollary

If $\{x_{n}\}_{n=1}^\infty$ is not Cauchy, then it does not converge.

Example

Let $z_{n} = (-1)^{n}$. Note that

\[z_{n} = \begin{cases} 1 \quad\text{if $n$ is even} \\ -1 \quad\text{if $n$ is odd} \end{cases}\]

Let $0 < \varepsilon <2$ and $N_{0} \in \mathbb{N}$. Take $n = 2N_{0} > N_{0}$ and $m = 2N_{0}+1>N_{0}$. Then $\lvert x_{n} - x_{m}\rvert = 2 > \varepsilon$. Therefore $\{z_{n}\}_{n=1}^\infty$ is not Cauchy and diverges.

Definition (Divergence to infinity)

Let $\{x_{n}\}_{n=1}^\infty$ be a sequence. We say that it diverges to infinity, denoted by $x_{n} \underset{n \rightarrow \infty}{\longrightarrow} \infty$ ($\lim_{ n \to \infty } x_{n} = \infty$), if for every $\alpha > 0$ there exists $N \in \mathbb{N}$ such that $x_{n} > \alpha$ for every $n \geq N$. In the same way, $x_{n} \underset{n \rightarrow \infty}{\longrightarrow} -\infty$ if for every $\beta >0$ there exists $N \in \mathbb{N}$ such that $x_{n} < -\beta$ for every $n \geq N$.

Example

If $b>1$ and $y_{n} = b^{n}$, then $y_{n} \underset{n \rightarrow \infty}{\longrightarrow}\infty$. $\underset{}{\sup}$

Theorem (Monotone convergence)

Let $\{x_{n}\}_{n=1}^\infty \subseteq \mathbb{R}$.

If $\{x_{n}\}_{n=1}^\infty$ is decreasing and bounded below, then $x_{n} \underset{n \rightarrow \infty}{\longrightarrow} \underset{n \in \mathbb{N}}{\inf} x_{n}$.
If $\{x_{n}\}_{n=1}^\infty$ is increasing and bounded above, then $x_{n} \underset{n \rightarrow \infty}{\longrightarrow} \underset{n \in \mathbb{N}}{\sup} x_{n}$.

Proof of 2: Assume that $\{x_{n}\}_{n=1}^\infty$ is increasing and bounded above. Take $L = \underset{n \in \mathbb{N}}{\sup} x_{n}$ and $\varepsilon >0$ fixed and arbitrary. By the definition of the supremum, there exists $N \in \mathbb{N}$ such that $L - \varepsilon < x_{N} \leq L$. Take $n \geq N$, then $x_{n} \geq x_{N}$. Hence $L - \varepsilon < x_{N} \leq x_{n} \leq L$. Therefore, conclude that $x_{n} \underset{n \rightarrow \infty}{\longrightarrow}L$. The proof of 1. is analogous.

Example

Consider $z_{n} = a^{n}$ with $0<a<1$. Note that $a < 1$, so $a^{n+1} < a^{n}$. Let $L = \underset{n \in \mathbb{N}}{\inf} z_{n}$. Take $\varepsilon>0$. By the definition of the infimum, we know that there exists $N_{1} \in \mathbb{N}$ such that $L \leq a^{N_{1}} < L + \varepsilon$. Let $n \geq N_{1}$. Then $L \leq a^{n} < a^{N_{1}} < L + \varepsilon$, i.e. $\lvert a^{n}-L \rvert < \varepsilon$. Therefore $z_{n} \underset{n \rightarrow \infty}{\longrightarrow}L$. We will now prove that $L = 0$. To that end, suppose for the sake of contradiction that $L>0$, that is, that for every $n \in \mathbb{N}$, $0 < L < a^{n}$. Then $\left( \frac{1}{a} \right)^{n} < \frac{1}{L}$, a contradiction since $\frac{1}{a} > 1$ and in that case the sequence diverges to $\infty$.

Theorem (Algebraic properties of the limit)

Let $\{x_{n}\}_{n=1}^\infty$ and $\{y_{n}\}_{n=1}^\infty$ be sequences in $\mathbb{R}$ such that $x_{n} \underset{n \rightarrow \infty}{\longrightarrow}\ell_{1}$ and $y_{n} \underset{n \rightarrow \infty}{\longrightarrow}\ell_{2}$. Then:

$x_{n} + y_{n} \underset{n \rightarrow \infty}{\longrightarrow} \ell_{1} + \ell_{2}$.
$x_{n}y_{n} \underset{ \rightarrow \infty}{\longrightarrow} \ell_{1} \ell_{2}$.
If $y_{n} \neq 0$ for every $n \in \mathbb{N}$ and $\ell_{2} \neq 0$, then $\frac{1}{y_{n}} \underset{n \rightarrow \infty}{\longrightarrow} \frac{1}{\ell_{2}}$.

Proof of 3: We know that $\ell_{2} \neq 0$. Let $\varepsilon > 0$ be fixed and arbitrary. Then there exists $N_{1} \in \mathbb{N}$ such that

\[\begin{aligned} &\lvert y_{n} \rvert - \lvert \ell_{2} \rvert \leq \lvert y_{n}-\ell_2 \rvert < \varepsilon \leq \\ \implies & \lvert l_{2}\rvert - \varepsilon < \lvert y_{n} \rvert. \end{aligned}\]

I take $\varepsilon = \frac{\lvert l_{2} \rvert}{2}$, so $\lvert y_{n} \rvert \geq\lvert l_{2} \rvert - \varepsilon \geq \lvert \ell_{2} \rvert - \frac{\lvert l_{2} \rvert}{2} = \frac{\lvert \ell_{2} \rvert}{2}$ for $n \geq N_{1}$, that is, $\frac{1}{\lvert y_{n} \rvert} \leq \frac{2}{\lvert \ell_{2} \rvert}$. Then, for $n\geq N_{1}$, we have that

\[\left\lvert \frac{1}{y_{n}} - \frac{1}{\ell_{2}} \right\rvert = \frac{\lvert \ell_{2} - y_{n} \rvert }{\lvert y_{n} \rvert \lvert \ell_{2} \rvert } \leq \frac{\lvert \ell_{2}-y_{n} \rvert }{\lvert \ell_{2} \rvert} \frac{2}{\lvert \ell_{2} \rvert }.\]

Let $N_{2} \in \mathbb{N}$ be such that $\lvert y_{n} - \ell_{2} \rvert < \frac{\varepsilon \lvert \ell_{2} \rvert}{2}$. Then, for $n \geq \max\{N_{1}, N_{2}\}$, we have that

\[\left\lvert \frac{1}{y_{n}} - \frac{1}{\ell_{2}} \right\rvert = \frac{\lvert y_{n}-\ell_{2} \rvert }{\lvert y_{n} \rvert \lvert \ell_{2} \rvert } \leq \frac{2\lvert y_{n}-\ell_{2} \rvert }{\lvert \ell_{2} \rvert ^{2}} < \frac{\varepsilon \lvert \ell_{2} \rvert ^{2}}{2} \frac{2}{\lvert \ell_{2} \rvert ^{2}} = \varepsilon.\]

Conclude that $\frac{1}{y_{n}} \underset{n \rightarrow \infty}{\longrightarrow} \frac{1}{\ell_{2}}$.

Theorem (Limits and continuous functions)

Let $\{x_{n}\}_{n=1}^\infty$ be such that $x_{n} \underset{n \rightarrow \infty}{\longrightarrow} L$ and let $f:(a,b) \to \mathbb{R}$ be continuous with $L \in (a,b)$. Then $f(x_{n}) \underset{n \rightarrow \infty}{\longrightarrow} f(L)$.

Proof: Given $\varepsilon>0$, there exists $\delta > 0$ such that if $\lvert x-L \rvert < \delta$ then $\lvert f(x) - f(L) \rvert < \varepsilon$. Since $x_{n} \underset{n \rightarrow \infty}{\longrightarrow} L$, there exists $N_{0}$ such that $\lvert x_{n} - L\rvert < \delta$ if $n \geq N_{0}$. Hence, for every $n \geq N_{0}$, $\lvert x_{n} - L \rvert < \delta$ and consequently $\lvert f(x_{n}) - f(L) \rvert < \varepsilon$.

Example

$\sqrt{ 1+\frac{2}{1+\frac{1}{n}}} \underset{n \rightarrow \infty}{\longrightarrow} \sqrt{ 3 }$. Note that:

\[\frac{1}{n} \underset{n \rightarrow \infty}{\longrightarrow} 0\] \[\implies 1+\frac{1}{n} \underset{n \rightarrow \infty}{\longrightarrow} 1\]

since $1+x$ is continuous on $\mathbb{R}$,

\[\implies \frac{2}{1+\frac{1}{n}} \underset{n \rightarrow \infty}{\longrightarrow} 2\]

since $\frac{2}{x}$ is continuous on $(0,+\infty)$,

\[\implies \sqrt{ 1 + \frac{2}{1+\frac{1}{n}} } \underset{n \rightarrow \infty}{\longrightarrow} \sqrt{ 3 },\]

since $\sqrt{ 1+x }$ is continuous on $(0,+\infty)$.

Example

Compute $\lim_{ n \to \infty } k - \sqrt{ k^{2}-k }$.

Note that

\[\begin{aligned} x_{k} &= \frac{(k-\sqrt{ k^{2}-k })(k+\sqrt{ k^{2}-k })}{k + \sqrt{ k^{2}-k }} \\ &= \frac{k^{2} - (\sqrt{ k^{2}-k })^{2}}{k + \sqrt{ k^{2}-k }} \\ &= \frac{k}{k+\sqrt{ k^{2}-k }} \\ &= \frac{1}{1 + \sqrt{ 1-\frac{1}{k}}}. \end{aligned}\]

Then

\[x_{k} = \frac{1}{1+\sqrt{ 1-\frac{1}{k} }} \underset{n \rightarrow \infty}{\longrightarrow} \frac{1}{2}.\]

Example

Compute $\lim_{ n \to \infty } \ln(n^{2}-1) - \ln(n(n-1))$.

Note that

\[\begin{aligned} \ln(n^{2}-1) - \ln(n(n-1)) &= \ln\left(\frac{n^{2}-1}{n(n-1)}\right) \\ &= \ln\left( \frac{(n+1)(n-1)}{n(n-1)} \right) \\ &= \ln\left( \frac{n+1}{n} \right) \\ &= \ln\left( 1 + \frac{1}{n} \right) \underset{n \rightarrow \infty}{\longrightarrow} \ln(1) = 0. \end{aligned}\]

Recursive sequences

Example

Let $\{a_{n}\}_{n=1}^\infty$ be such that $a_{0} = 1$ and $a_{n+1} = \sqrt{ 1+a_{n} }$. Prove that it converges and compute its limit.

Proof: First, we will prove by induction that $\{a_{n}\}_{n=1}^\infty$ is increasing, i.e., $a_{n+1} \geq a_{n}$ for every $n \in \mathbb{N}$. Base case: For $n = 0$, $a_{1} = \sqrt{ 2 } \geq 1 = a_{0}$. This proves the base case. Inductive step: Suppose as the inductive hypothesis that $a_{m+1} \geq a_{m}$ for some fixed and arbitrary $m \in \mathbb{N}$. We must show that $a_{m+2} \geq a_{m+1}$. From the inductive hypothesis:

\[\begin{aligned} a_{m+1} &\geq a_{m} \\ 1 + a_{m+1} &\geq 1+a_{m} \\ a_{m+2} = \sqrt{ 1 + a_{m+1} } &\geq \sqrt{ 1 + a_{m} } = a_{m+1}. \end{aligned}\]

This proves the inductive step. Conclude that $\{a_{n}\}_{n=1}^\infty$ is increasing. We will now prove that it is bounded above. In particular, we will prove by induction that for every $n \in \mathbb{N}$, $a_{n} \leq 2$. Base case: For $n=0$, $a_{0} = 1 \leq 2$. This proves the base case. Inductive step: Suppose as the inductive hypothesis that $a_{m} \leq 2$ for some fixed and arbitrary $m \in \mathbb{N}$. We must show that $a_{m+1} \leq 2$. From the inductive hypothesis:

\[\begin{aligned} a_{m} &\leq 2 \\ 1 + a_{m} &\leq 3\\ a_{m+1} = \sqrt{ 1 + a_{m} } &\leq \sqrt{3} < 2. \end{aligned}\]

This proves the inductive step. Conclude that $\{a_{n}\}_{n=1}^\infty$ is bounded. Then, by the monotone convergence theorem, conclude that $\{a_{n}\}_{n=1}^\infty$ converges. Let $L = \lim_{ n \to \infty } a_{n}$. Then

\[\begin{aligned} a_{n+1} &= \sqrt{ 1 + a_{n} } \\ \implies \lim_{ n \to \infty } a_{n+1} &=\lim_{ n \to \infty } \sqrt{1 + a_{n}} \\ \implies L &= \sqrt{ 1 + L } \\ \implies L^{2}-L-1 &= 0 \\ \implies L &= \frac{1+\sqrt{ 5 }}{2}. \end{aligned}\]

Example

Let $x_{n} = \frac{x_{n-1}+1}{3} = \frac{x_{n-1}}{3} + \frac{1}{3}$ for $n \geq 1$ and $x_{0} = x$. Note that

\[\begin{aligned} x_{1} &= \frac{x}{3}+\frac{1}{3} \\ x_{2} &= \frac{ \frac{x+1}{3}}{3} + \frac{1}{3} = \frac{x}{3^{2}} + \frac{1}{3^{2}} +\frac{1}{3} \\ x_{3} &= \frac{x}{3^{3}} + \frac{1}{3^{3}} + \frac{1}{3^{2}} + \frac{1}{3}. \end{aligned}\]

In general, one can show by induction that

\[x_{n} = \frac{x}{3^{n}} + \frac{1}{3^{n}} + \frac{1}{3^{n-1}} + \dots + \frac{1}{3}.\]

Now, if $a \neq 1$, $1+a+a^{2}+\dots+a^{n} = \frac{1-a^{n+1}}{1-a}$. Then

\[x_{n} = \frac{x}{3^{n}} + \frac{ 1-\left( \frac{1}{3} \right)^{n+1} }{\frac{2}{3}} - 1 = \frac{x}{3^{n}} + \frac{3}{2}\left( 1-\left( \frac{1}{3} \right)^{n+1} \right) -1.\]

Therefore $x_{n} \underset{n \rightarrow \infty}{\longrightarrow} \frac{1}{2}$.

Theorem (Convergence under permutation)

Let $\phi: \mathbb{N} \to\mathbb{N}$ be a bijection and let $\{a_{n}\}_{n=0}^\infty$ be such that $\sum_{n=0}^\infty \lvert a_{n} \rvert$ converges. Then $\sum_{n=0}^\infty a_{\phi(n)}$ converges and moreover $\sum_{n=0}^\infty a_{\phi_{n}} = \sum_{n=0}^\infty a_{n}$.

Proof: Define

\[S_{k} = \sum_{n=0}^{k} a_{n}, \quad \tilde{S}_{k} = \sum_{n=0}^{k} \lvert a_{n} \rvert, \quad U_{k} = \sum_{n=0}^{k} a_{\phi(n)}.\]

We know that $\sum_{n=0}^\infty \lvert a_{n} \rvert$ converges. Then, given $\varepsilon>0$, there exists $N$ such that $\lvert \tilde{S}_{n} - \tilde{S}_{m} \rvert < \varepsilon$ for all $m\geq n\geq N$. Then

\[\lvert a_{n+1} \rvert + \lvert a_{n+2} \rvert +\dots+ \lvert a_{m} \rvert < \varepsilon.\]

We must show that $\lim_{ k \to \infty } S_{k} = \ell = \lim_{ k \to \infty } U_{k}$. Take $N_{1}$ such that $\lvert S_{k} - \ell \rvert < \varepsilon$ if $k \geq N_{1}$. Since $\phi$ is a bijection, we know that for every $m \in \mathbb{N}$ there exists $n_{m}$ such that $\phi(n_{m})$. Take $M = \max \phi^{-1}[\{ 1,\dots,N \}]$. Then $\{ 1,\dots,N \} \subseteq \{ \phi(1),\dots,\phi(M) \}$. Then, for $m\geq M$, $k \geq N$, we have that

\[\lvert U_{m} - S_{k} \rvert \leq \lvert a_{N+1} \rvert +\dots+\lvert a_{\ell} \rvert < \varepsilon.\]