Proposition 1.Let $a:\mathbb{N}\to\mathbb{R}^+$ be an unbounded and strictly increasing sequence such that $\lim_{n\to\infty} a_n/a_{n+1}=1$. Then
\[M := \left\{\frac{a_n}{a_m} : n,m\in\mathbb{N}\right\}\]is a dense subset of $\mathbb{R}^+$, meaning that for all $x$, $y\in\mathbb{R}^+$ with $x<y$ there is a $z\in M$ such that $x<z<y$.
Proof.Let $x$, $y\in\mathbb{R}^+$ with $x<y$. Let $\varepsilon = 1-x/y$, then $\varepsilon>0$. Since $\lim_{n\to\infty} a_n/a_{n+1}=1$, there is an $N\in\mathbb{N}$ such that
\[\forall n\geq N : \left\lvert\frac{a_n}{a_{n+1}}-1\right\rvert < \varepsilon.\]Since $a$ is strictly increasing, $a_n/a_{n+1}<1$ for all $n\in\mathbb{N}$. So
\[\forall n\geq N : 1 - \frac{a_n}{a_{n+1}} < \varepsilon.\]We have
\[\forall n\geq N : \frac{a_n}{a_{n+1}} > 1 - \varepsilon = 1 - \left(1 - \frac{x}{y}\right) = \frac{x}{y}. \tag{$\ast$}\]Since $a$ is unbounded, there is an $m\in\mathbb{N}$ such that
\[a_m \geq \frac{a_N}{x} \quad\Rightarrow\quad a_m x \geq a_N.\]We assert there is an $n\in\mathbb{N}$ such that $a_m x < a_n < a_m y$. Assume the opposite. Let
\[k = \max\{i\in\mathbb{N} : a_i \leq a_m x\}.\](this maximum exists - since $a$ is unbounded and strictly increasing, $\{i\in\mathbb{N} : a_i \leq a_m x\}$ can not be infinite), then $a_k\leq a_m x$ and $a_{k+1}\geq a_m y$. It follows
\[\frac{a_k}{a_{k+1}} \leq \frac{a_m x}{a_m y} = \frac{x}{y}.\]On the other hand, since $a_N\leq a_m x$, $k\geq N$. Hence by $(\ast)$ we have
\[\frac{a_k}{a_{k+1}} > \frac{x}{y},\]a contradiction. Therefore, there do exist $n$, $m\in\mathbb{N}$ such that
\[x < \frac{a_n}{a_m} < y.\]$\square$
Lemma 1.Suppose $A$ and $B$ are two disjoint sets. If $A$ is countably infinite and $B$ is countable, then $A\cup B$ is countably infinite.
Proof.Since $A$ is countably infinite, there is a bijection $f:\mathbb{N}\to A$.
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In case $B$ is finite, $\lvert B\rvert\in\mathbb{N}$ and there is a bijection $g:\lvert B\rvert\to B$. Define
\[h : \mathbb{N} \to A\cup B,\ n \mapsto \begin{cases} f(n), & \text{if $n\geq\lvert B\rvert$;} \\ g(n), & \text{otherwise.} \end{cases}\]Then it can be verified that $h$ is bijective.
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In case $B$ is infinite, there is a bijection $g:\mathbb{N}\to B$. Define
\[h : \mathbb{N} \to A\cup B,\ n \mapsto \begin{cases} f(n), & \text{if $n$ is even;} \\ g(n), & \text{otherwise.} \end{cases}\]Then it can be verified that $h$ is bijective. $\square$
Proposition 2.Suppose $A$ is an infinite set, $B\subseteq A$ is a countable subset such that $A\setminus B$ is infinite. Then $A\setminus B$ is equinumerous to $A$.
Proof.Since $A\setminus B$ is infinite, there is a countably infinite subset $C\subseteq A\setminus B$. By Lemma 1, there is a bijection $f:B\cup C\to C$. Define
\[g : A \to A\setminus B,\ x \mapsto \begin{cases} f(x), & \text{if $x\in B\cup C$;} \\ x, & \text{otherwise.} \end{cases}\]Then it can be verified that $g$ is bijective. $\square$
Proposition 3.The set of all increasing sequences in $\mathbb{N}$ is uncountable.
Proof (diagonal argument).Denote the set of all increasing sequences in $\mathbb{N}$ by $I$. Assume $I$ is countable, then there is a bijection $f:\mathbb{N}\to I$. Define the sequence $a:\mathbb{N}\to\mathbb{N}$ recursively by
\[a_0 = f(0)_0 + 1; \quad a_n = \max\{a_{n-1},f(n)_n\} + 1\ \text{for all $n\geq 1$.}\]For every $n\geq 1$, $a_n\geq a_{n-1}+1>a_{n-1}$, which means that $a$ is (strictly) increasing, so $a\in I$. Since $f$ is bijective, there is an $m\in\mathbb{N}$ such that $f(m)=a$. Since $a_0=f(0)_0+1$, $b_0\neq f(0)_0$, so $a\neq f(0)$. It means $m>0$ and as a result we have
\[a_m = \max\{a_{m-1},f(m)_m\} + 1 \geq f(m)_m + 1 = a_m + 1,\]a contradiction. $\square$
Remark 1.As an exercise for the reader: $\lvert I\rvert=2^{\aleph_0}$.
Proposition 4.The set of all decreasing sequences in $\mathbb{N}$ is countable.
Proof.It can be observed that every decreasing sequence in $\mathbb{N}$ can be encoded by a finite sequence in $\mathbb{N}$, and the set of all finite sequences in $\mathbb{N}$ is countable. One way to think about this is to use the Fundamental Theorem of Arithmetic: Let $S$ be set of all finite sequences in $\mathbb{N}$. Define
\[f : S \to \mathbb{N},\ (a_0,\cdots,a_k) \mapsto \prod_{i=0}^{k}p_{i+1}^{a_i+1},\]where $p_n$ ($n\geq 1$) is the $n$-te prime number. Then $f$ is injective. $\square$
For another proof see this answer on StackExchange.
Proposition 5.Let $x$, $y:\mathbb{N}\to\mathbb{R}$ be two bounded sequences.
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$\limsup_{n\to\infty} (x_n + y_n) \leq \limsup_{n\to\infty} x_n + \limsup_{n\to\infty} y_n$.
Proof.Since $x$ and $y$ are bounded, $x+y$ is bounded. So $\limsup_{n\to\infty} x_n$, $\limsup_{n\to\infty} y_n$ and $\limsup_{n\to\infty} (x_n + y_n)$ exist. There are subsequences $x\circ T$, $y\circ\tilde{T}$ and $(x+y)\circ\tau$ such that
\[\limsup_{n\to\infty} x_n = \lim_{n\to\infty} x_{T(n)}, \quad \limsup_{n\to\infty} y_n = \lim_{n\to\infty} y_{\tilde{T}(n)}\]and
\[\limsup_{n\to\infty} (x_n + y_n) = \lim_{n\to\infty} (x_{\tau(n)}+y_{\tau(n)}).\]Since $x$ is bounded, $x\circ\tau$ is bounded. By Bolzano-Weierstrass Theorem, there is a convergent subsequence $x\circ\tau\circ\alpha$. Since $(x+y)\circ\tau\circ\alpha$ is a subsequence of $(x+y)\circ\tau$, we have
\[\lim_{n\to\infty} (x_{\tau(\alpha(n))} + y_{\tau(\alpha(n))}) = \lim_{n\to\infty} (x_{\tau(n)} + y_{\tau(n)}) = \limsup_{n\to\infty} (x_n + y_n).\]From this we also see $y\circ\tau\circ\alpha$ is convergent. Since $\limsup$ is the largest accumulation point, we have
\[\lim_{n\to\infty} x_{\tau(\alpha(n))} \leq \lim_{n\to\infty} x_{T(n)} \quad\text{und}\quad \lim_{n\to\infty} y_{\tau(\alpha(n))} \leq \lim_{n\to\infty} y_{\tilde{T}(n)}.\]Therefore,
\[\begin{align*} \limsup_{n\to\infty} (x_n + y_n) &= \lim_{n\to\infty} (x_{\tau(\alpha(n))} + y_{\tau(\alpha(n))}) = \lim_{n\to\infty} x_{\tau(\alpha(n))} + \lim_{n\to\infty} y_{\tau(\alpha(n))} \\ &\leq \lim_{n\to\infty} x_{T(n)} + \lim_{n\to\infty} y_{\tilde{T}(n)} = \limsup_{n\to\infty} x_n + \limsup_{n\to\infty} y_n. \end{align*}\]$\square$
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If $x$ converges, then $\limsup_{n\to\infty} (x_n + y_n) = \limsup_{n\to\infty} x_n + \limsup_{n\to\infty} y_n$.
Proof.Since “$\leq$” is shown in 1., we only have to show “$\geq$”. Let $\tau$, $\alpha$, $T$ and $\tilde{T}$ defined as in 1. Since $x$ converges, all subsequences of $x$ have the same limit. Therefore,
\[\begin{align*} \limsup_{n\to\infty} x_n + \limsup_{n\to\infty} y_n &= \lim_{n\to\infty} x_{T(n)} + \lim_{n\to\infty} y_{\tilde{T}(n)} = \lim_{n\to\infty} x_{\tilde{T}(n)} + \lim_{n\to\infty} y_{\tilde{T}(n)} \\ &= \lim_{n\to\infty} (x_{\tilde{T}(n)} + y_{\tilde{T}(n)}) \leq \limsup_{n\to\infty} (x_n + y_n). \end{align*}\]$\square$