# Convergent & Cauchy Sequence related prove

(1) Consider the two convergent sequences $\{a_n\}$and $\{b_n\}$ such that $$\{a_n\}\to a$$ and $$\{b_n\}\to b$$ for $n\to\infty$.

Prove that $$\{a_n+b_n\}\to a + b$$ for $n\to\infty$.

(2) Prove that a convergent sequence is Cauchy. (Recall also that a Cauchy sequene is a convergent sequence).

• These are results that would appear in any introductory analysis textbook. Do you have a specific question about the proof? – Ittay Weiss Apr 1 '13 at 6:06
• I suppose you only work in the context of real analysis, because what's said between brackets is not true generally. – Raskolnikov Apr 1 '13 at 6:08

1) Let $\varepsilon > 0$ then there exist $N \in \mathbb{N}$ such that $|a_n - a| < \varepsilon/2$ and $|b_n - b| < \varepsilon/2$ for $n \geq N$. Thus, by the triangle inequality we have $$|(a_n + b_n) - (a+b) | \leq | a_n - a | + |b_n - b| < \varepsilon/2 + \varepsilon/2 = \varepsilon$$ Hence $(a_n + b_n) \rightarrow (a+b)$
2) Suppose $a_n \rightarrow a$. Then $\forall \; \varepsilon > 0$ then there exist $N \in \mathbb{N}$ such that $|a_n - a| < \varepsilon/2$ for $n \geq N$. Hence for $m\geq n \geq N$ we have $$|a_n - a_m | = |(a_n - a) + (a-a_m) | \leq |(a_n - a)| + |(a-a_m) | < \varepsilon/2 + \varepsilon/2 = \varepsilon$$ Hence $a_n$ is Cauchy.