0
$\begingroup$

This question already has an answer here:

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

I start with this hypothesis

$\displaystyle\lim_{n \to\infty}{a_n}=L \Leftrightarrow{\forall{\epsilon}>0}$ $\exists{N}\in{\mathbb{N}}$ 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!!!

$\endgroup$

marked as duplicate by dfeuer, user61527, Hanul Jeon, Norbert, Stefan4024 Oct 27 '13 at 8:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 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. $\endgroup$ – user99914 Oct 27 '13 at 4:14
  • $\begingroup$ In a nutshell, use triangle inequality and choose your epsilons appropriately $\endgroup$ – Hawk Oct 27 '13 at 4:15
  • $\begingroup$ Note: the question in the (approximate) duplicate is rather terrible. Don't read it. Just read the answer. $\endgroup$ – dfeuer Oct 27 '13 at 4:25
  • $\begingroup$ Another: math.stackexchange.com/questions/325836/… $\endgroup$ – dfeuer Oct 27 '13 at 4:27
0
$\begingroup$

$$ \textbf{Solution} $$

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.