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It was a question of my integral calculus exam:

Write the Cauchy definition of $\lim_{n\rightarrow\infty} a_n = L$

So, I literally wrote:

" $\{a_n\}$ is a Cauchy sequence if given $\epsilon>0$ there exists $N \in \mathbb{N}$ such that $\forall\ m,n\in\mathbb{N}$ if $n,m\ge N$, then $|a_n - a_m|<\epsilon$, or with symbols:

$\forall\epsilon>0:\exists N\in\mathbb{N}:\forall\ m,n \in\mathbb{N}: m,n\ge N\implies |a_n - a_m|<\epsilon$.

And if $\{a_n\}$ is a Cauchy sequence, then $\lim_{n\rightarrow\infty} a_n = L$ ( i.e. $\{a_n\}$ converges) "

The following question was:

With this definition, find $\lim_{n\rightarrow\infty} (3n+2)$

And that is what I did:

I proposed that $\lim_{n\rightarrow\infty} (3n+2) = \infty$ i.e. $\{3n+2\}$ diverges. So I think that I have to prove that this limit is equal to infinity using the Cauchy definition.

But if a sequence $\{a_n\}$ diverges $\implies$ $\{a_n\}$ is not a Cauchy sequence.

And $\{a_n\}$ is not a Cauchy sequence if

$\exists\epsilon>0:\forall N\in\mathbb{N}:\exists\ m,n \in\mathbb{N}:$ $m,n\ge N$ and $|a_n - a_m|\ge\epsilon$ $(1)$

So, I must to prove $(1)$.

$(1)$ says that there exists $m,n\in\mathbb{N}$ such that ..., so $(1)$ has not to apply $\forall m,n\in\mathbb{N}$, and for this fact (I think) I can assign convinient values to $m,n$ in order to prove $(1)$:

If i choose $m=N+1,n=N$ then $|3(N+1)+2 -(3N+2)|=3>2$.

So exists $\epsilon=2>0 $ such that $\forall N \in\mathbb{N}:$ there exists $m=N+1, n=N$ such that $N+1,N \ge N$ and $|3(N+1)+2 -(3N+2)|=3>\epsilon$.

Hence $\{3n+2\}$ is not a Cauchy sequence $\implies$ $\lim_{n\rightarrow\infty} (3n+2) = \infty$.

Is my proof correct? If it is not correct, could you help me proving that, please?

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  • $\begingroup$ just reading quick, it looks perfectly good to me $\endgroup$ – Chessnerd321 Jun 28 '17 at 19:57
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    $\begingroup$ I will note that, at the end, we cannot just say that $$\text{$\{a_n\}$ not Cauchy$\implies \lim_{n\to\infty}a_n=\infty$}$$We can only say that $$\text{$\{a_n\}$ not Cauchy$\implies \lim_{n\to\infty}a_n\notin\Bbb R$}$$The limit may be $\pm\infty$ or just non-existent. $\endgroup$ – Dave Jun 28 '17 at 20:00
  • $\begingroup$ But if i i want to prove that the limit is $ \infty$, what have i to do? $\endgroup$ – Sama Jun 28 '17 at 20:05
  • $\begingroup$ @KarenSM You have to show that for each $M >0$, there is a natural number $N$ such that for all $n \geq N$, $a_n\geq M$. Quite trivial, you can take $N=[M]$, where $[M]$ is the “floor” of $M$, the unique integer satisfying $M-1< [M] \leq M$. $\endgroup$ – Li Chun Min Jun 28 '17 at 23:53
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    $\begingroup$ @LiChunMin as far as I understand you, I cannot find the $\lim_{n\rightarrow\infty} (3n+2) $ with the Cauchy definition, right? $\endgroup$ – Sama Jun 29 '17 at 0:26
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The definition of a Cauchy sequence is:

$\forall \epsilon > 0, \exists N \in \mathbb{N}: \forall m,n \in \mathbb{N}: m,n \ge N \implies |a_m-a_n| < \epsilon$, and the negation of this definition is:

$\exists \epsilon > 0, \forall N \in \mathbb{N}, \exists m,n \in \mathbb{N}: m,n \ge N, |a_m - a_n| \ge \epsilon$.

For your sequence $a_n = 3n+2, n \ge 1$, choose $\epsilon = 1$ (your choice is $3$), for any $N \in \mathbb{N}$, let $n =N, m = N+1$. Observe $m, n \ge N$, and $|a_m - a_n| = |a_{N+1} - a_{N}|= |3(N+1) - 2 - 3N+2|= 3 > 1= \epsilon $. Thus the sequence above is not Cauchy.

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  • $\begingroup$ But with $\epsilon=2$ is still fine, right? I'm nervous because I put this in my exam $\endgroup$ – Sama Jun 28 '17 at 20:22
  • $\begingroup$ $\epsilon = 2$ works just fine. $2$ is better than $1$... $\endgroup$ – DeepSea Jun 28 '17 at 20:44
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You did not write the def'n of $\lim_{n\to \infty}a_n=L.$ What you wrote in response to "Define what it means for $(a_n)_n$ to converge to $L$" was: " $(a_n)_n$ is a Cauchy sequence (which means ...) which converges to $L$."

The def'n of $\lim_{n\to \infty}a_n=L$ is $$\forall r>0\;\exists m\;\forall n>m\;(|L-a_n|<r).$$ I have omitted specifying that $m,n\in \mathbb N.$ It is conventional that in "$\lim_{n\to \infty}a_n$", the values of $n$ are restricted to members of $\mathbb N$ unless stated otherwise.

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