# Proof that $\lim{a_n}=L$ when $n$ goes to infinity, then $\{a_n\}$ its a Cauchy Sequence [duplicate]

Proof that $\lim{a_n}=L$ when $n$ goes to infinity, then $\{a_n\}$ its a Cauchy Sequence

$\displaystyle\lim_{n \to\infty}{a_n}=L \Leftrightarrow{\forall{\epsilon}>0}$ $\exists{N}\in{\mathbb{N}}$ such that $\forall{n}>N \left |{a_n-L}\right |< \epsilon$

The same its true fot any $m>N,\left |{a_m-L}\right |=\left |{L-a_m}\right |< \epsilon$

I will sum the inequalities but I havent any idea how do it... Help me please!!!

## marked as duplicate by dfeuer, user61527, Hanul Jeon, Norbert, Stefan4024Oct 27 '13 at 8:09

• Hint: $|a_m - a_n| \leq |a_n - L| + |a_m - L|$. It just say that if $a_n$ and $a_m$ are close to $L$, then $a_n$ and $a_m$ are close to each other. – user99914 Oct 27 '13 at 4:14
• In a nutshell, use triangle inequality and choose your epsilons appropriately – Hawk Oct 27 '13 at 4:15
• Note: the question in the (approximate) duplicate is rather terrible. Don't read it. Just read the answer. – dfeuer Oct 27 '13 at 4:25
• – dfeuer Oct 27 '13 at 4:27

$$\textbf{Solution}$$

Let $\epsilon > 0$ be given. Suppose $\lim a_n = L$. Therefore, we can take $N_1, N_2$ such that

$$|a_n - L | < \epsilon/2 \; \; \; \text{for all} \;\;n \geq N_1$$

$$|a_m - L | < \epsilon/2 \; \; \; \text{for all} \;\;m \geq N_2$$

We can find such $N_1$ and $N_2$ by the definition of limit. Now, Take $N = \max\{N_1, N_2 \}$

$$\therefore |a_n - a_m| = |a_n - L + L - a_m| \leq |a_n - L| + |L - a_m| < \epsilon/2 + \epsilon/2 = \epsilon$$

For every $m, n \geq N$. Note the triangle trick have been used in the above estimate.

Therefore, the sequence must be Cauchy and the problem is solved