Prove that sequences with finite limits are linear with respect to taking limits

Let

$\lim_{n \to \infty} a_n = A \qquad \text{and} \qquad \lim_{n \to \infty} b_n = B.$

Prove directly from the definition of the limit that

$\lim_{n \to \infty} (a_n + b_n) = A + B \qquad \text{and} \qquad \lim_{n \to \infty} (ca_n) = cA$

for any constant $c$ .

Proof. Let $\varepsilon >0$ be given. Since $\lim_{n \to \infty} a_n = A$ and $\lim_{n \to \infty} b_n = B$ we know there exist positive integers $N_a$ and $N_b$ such that

$|a_n - A| < \frac{\varepsilon}{2} \qquad \text{and} \qquad |b_m - B| < \frac{\varepsilon}{2}$

for all $n \geq N_a$ and all $m \geq N_b$ . Let $N = \max \{ N_a, N_b \}$ . Then, for all $n \geq N$ we have

$\begin{align*} &&|a_n - A| + |b_n - B| &< \varepsilon \\ \implies && |a_n + b_n - (A+B)| &< \varepsilon. \end{align*}$

Hence,

$\lim_{n \to \infty} (a_n + b_n) = A+B. \qquad \blacksquare$

Proof. Let $\varepsilon > 0$ be given. Since $\lim_{n \to \infty} a_n = A$ we know there exists a positive integer $N$ such that

$|a_n - A| < \frac{\varepsilon}{|c|} \qquad \text{for all } n \geq N.$

Thus,

$\begin{align*} &&|c| \cdot |a_n - A| &< |c| \cdot \frac{\varepsilon}{|c|} \\[9pt] \implies && | ca_n - cA| &< \varepsilon \\[9pt] \implies && \lim_{n \to \infyt} ca_n &= cA. \qquad \blacksquare \end{align*}$

One comment

November 9, 2019 at 12:12 pm

Artem says:

There is a small problem in the second proof: you cannot just divide by |c|, since it can be 0. Thus, there must be 2 cases: case |c| > 0, and c = 0.

Stumbling Robot

Prove that sequences with finite limits are linear with respect to taking limits

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