📂Analysis

Properties of Converging Real Sequences

Theorem 1[^1]

Let $\left\{ s_{n} \right\}$, $\left\{ t_{n} \right\}$ be sequences of real (or complex) numbers and assume that $\lim \limits_{n\to\infty} s_{n}=s$, $\lim\limits_{n\to\infty}t_{n}=t$. Then

• (a) $\lim \limits_{n\to\infty}(s_{n}+t_{n})=s+t$

• (b) $\forall c \in \mathbb{C},\quad\lim \limits_{n\to\infty} cs_{n}=cs \quad \text{and} \quad \lim \limits_{n\to\infty} (c+s_{n})=c+s$

• (c) $\lim \limits_{n\to\infty} s_{n}t_{n}=st$

• (d) $\forall s_{n}\ne 0,s\ne0,\quad \lim \limits_{n\to\infty}\frac{1}{s_{n}}=\frac{1}{s}$

---

Of course, this can be extended to $\mathbb{R}^{k}$ as well. See Theorem 2 for a deeper look.

Proof

(a)

Given any positive number $\varepsilon>0$, there exist two positive numbers $N_{1}$, $N_{2}$ satisfying the following condition.

$$\begin{align*} n \ge N_{1} &\implies \left|s_{n}-s \right|<\frac{\varepsilon}{2} \\ n \ge N_{2} & \implies \left|t_{n}-t \right|<\frac{\varepsilon}{2} \end{align*}$$

Now let’s assume that $N=\max(N_{1},N_{2})$. For $n \ge N$,

$$\left| (s_{n}+t_{n})-(s+t) \right| \le\left| s_{n}-s \right|+\left| t_{n} -t\right|<\varepsilon$$

Therefore,

$$\lim \limits_{n\to\infty} (s_{n}+t_{n})=s+t$$

■

(b)

This trivially holds due to the validity of (a).

■

$(c)$

Given any positive number $\varepsilon >0$, there exist two positive numbers $N_{1}$, $N_{2}$ that satisfy the following equation.

$$\begin{align*} n \ge N_{1} &\implies \left|s_{n}-s \right|<\sqrt{\varepsilon} \\ n \ge N_{2} & \implies \left|t_{n}-t \right|< \sqrt{\varepsilon} \end{align*}$$

Now let’s assume that $N=\max (N_{1},N_{2})$. Therefore,

$$n \ge N \implies \left| (s_{n}-s) (t_{n}-t) \right|<\varepsilon$$

Hence,

$$\begin{equation} \lim \limits_{n\to\infty} (s_{n}-s)(t_{n}-t)=0 \label{eq1} \end{equation}$$

At this moment,

$$s_{n}t_{n}-st=(s_{n} -s)(t_{n}-t)+s(t_{n}-t)+t(s_{n}-s)$$

Applying (a), (b), and $\eqref{eq1}$ to the above equation gives

$$\begin{align*} \lim \limits_{n\to\infty} (s_{n}t_{n}-st)&=\lim \limits_{n\to\infty}(s_{n}-s)(t_{n}-t)+\lim \limits_{n\to\infty}s(t_{n}-t)+\lim \limits_{n\to\infty}t(s_{n}-s) \\ &= 0+0+ 0 \\ &= 0 \end{align*}$$

Therefore,

$$\lim \limits_{n\to\infty} s_{n}t_{n}=st$$

■

(d)

Given the assumption of $\lim \limits_{n\to\infty} s_{n}=s$, one can choose a positive number $m$ that satisfies the following equation.

$$\forall n\ge m,\quad \left|s_{n}-s \right| < \frac{1}{2}\left|s \right|$$

Since $\left|s \right|-\left|s_{n} \right| \le \left|s_{n}-s \right|$, from the above equation, we can derive the following equation.

$$\begin{equation} \forall n \ge m,\quad \left|s_{n} \right|>\frac{1}{2}\left|s \right| \label{eq2} \end{equation}$$

Now, given any positive number $\varepsilon>0$, there exists a $N$ satisfying the equation below.

$$\begin{equation} n \ge N \implies \left|s_{n}-s \right| < \frac{1}{2}\left|s \right|^{2}\varepsilon \label{eq3} \end{equation}$$

Therefore, by $\eqref{eq2}$ and $\eqref{eq3}$, for $n \ge N$,

$$\left|\frac{1}{s_{n}}-\frac{1}{s} \right|=\left|\frac{s_{n}-s}{s_{n}s} \right|<\frac{2}{\left|s \right|^{2}}\left|s_{n}-s \right|<\varepsilon$$

thus,

$$\lim \limits_{n\to\infty}\frac{1}{s_{n}}=\frac{1}{s}$$

■