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I have been scratching my head for a couple of days on how to determine convergence/divergence of sequences. I made it to understand how to prove that a sequence converges, but still have numerous doubts about proof of divergence.

Say I have $\lim_{n \to +\infty } \sqrt{n+1} = +\infty$ and I have to prove the sequence diverges.

What I did is using the definition of converging sequence

$| a_n - L | < \ \varepsilon$

Where L is a theoretical limit (Fixed, real number) and $\varepsilon$ is also a theoretical, real number bounding the sequence (am I understanding this correctly?)

Then, I tried the proof by contradiction by doing

$-\varepsilon \ < \sqrt{n+1} - L < \varepsilon$

$ -\varepsilon + L < \sqrt{n+1} < \varepsilon + L$

$(-\varepsilon + L)^2 - 1 < n < (\varepsilon + L)^2 - 1$

Assuming that $\varepsilon$ and L are fixed, real numbers, we can always come up with an n greater than any operation done between those numbers, thus contradicting the fact that a bound exists.

Is the proof I've come up with valid and sufficient ?

I want to apologize in advance to people familiar with this, in case I made a horrible mess.

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  • $\begingroup$ This is hard to follow. What is $L$ in your example? $\endgroup$
    – lulu
    Jan 19 at 1:22
  • $\begingroup$ A theoretical real number that would be a limit of the sequence. So basically limit. $\endgroup$
    – Eugenio T.
    Jan 19 at 1:23
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    $\begingroup$ To show that $\lim_{n\to \infty} f(n)=\infty$ one must prove that, for all $M$, there exists $N=N(M)$ such that $n>N\implies f(n)>M$. Very different from what you wrote. $\endgroup$
    – lulu
    Jan 19 at 1:23
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    $\begingroup$ That would be proof that a sequence increases continually. I proved it by contradicting the definition of convergence. Or am I missing something and I cannot use proof by contradiction in this case? $\endgroup$
    – Eugenio T.
    Jan 19 at 1:26
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    $\begingroup$ My interpretation is that the OP knows that the limit "equals $\infty$" and is being asked to prove that the limit does not exist, that is, the limit does not equal any real number $L$. I think the argument is basically correct, although it could be written a bit more clearly. $\endgroup$ Jan 19 at 1:30

1 Answer 1

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There are often multiple ways to prove something, and both of the methods suggested in the comments are valid.

To describe and compare both methods, let me first review the definition of convergence.

To say that a sequence $a_n$ converges means:

There exists $L \in \mathbb R$ such that for every $\epsilon > 0$ there exists $N \in \mathbb N$ such that for every $n \in \mathbb N$, if $n \ge N$ then $|a_n - L| < 0$.

First method. The method you followed was argument by contradiction: Assume that $a_n = \sqrt{n+1}$ converges and argue to a contradiction. As suggested by @GregMartin, your proof is basically just fine, but you ought to have started it by saying explicitly that you are doing a proof by contradiction, and you ought to have expressed the logical steps more clearly, something like this:

  • Assume that $a_n = \sqrt{n+1}$ converges.
  • Applying the definition of convergence, choose $L \in \mathbb R$ as in that definition.
  • Therefore, for each $\epsilon > 0$ we know that there exists $N \in \mathbb N$ such that for each $n \in \mathbb N$, if $n \ge N$ then $|\sqrt{n+1}-L| < \epsilon$.

From this point, you manipulate that final inequality just as you did. Then, at the end, you very clearly state contradiction that you arrive at.

Second method. This method has two steps.

  • Step 1: Prove that $\sqrt{n+1}$ diverges to $+\infty$. To do that you must apply the definition: to say that that $a_n$ diverges to $+\infty$ means:

For every $M > 0$ there exists $N \in \mathbb N$ such that for every $n \in \mathbb N$, if $n \ge N$ then $a_n \ge M$.

  • Step 2: Prove a general lemma: if a sequence $a_n$ diverges to $+\infty$ then $a_n$ does not converge.

The proof of this general lemma is going to be an argument by contradiction (somewhat like what you did already): You assume that $a_n$ converges to some limit $L$, and you assume that $a_n$ diverges to $+\infty$, and then you argue to a contradiction.

To compare these, method 2 is more complicated that method 1, but method 2 has some important advantages: it produces more information about the sequence $a_n = \sqrt{n+1}$; and you prove a useful general lemma!

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  • $\begingroup$ So, a bit of a stupid question... But in the second method M and N are exactly what? Arbitrary numbers? If so, what is the link between them? $\endgroup$
    – Eugenio T.
    Jan 19 at 2:00
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    $\begingroup$ They are quantified variables. You already used the quantified variable $L$ in your proof that $\sqrt{n+1}$ does not converge. You also implicitly used the quantified variable $N$, when you spoke about "an $n$ greater than any operation done between those two numbers". $\endgroup$
    – Lee Mosher
    Jan 19 at 2:03
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    $\begingroup$ Mastering the clear expression of quantified variables, and the correct logical use of such variables, is one of the skills that makes a rock solid, valid proof. $\endgroup$
    – Lee Mosher
    Jan 19 at 2:04
  • $\begingroup$ So to put it in very basic terms and visualizing it on a cartesian plane, $M$ is going to be the boundary referring to points on $y$ axis, and $N$ is a boundary referring to $x$ axis? At least sort of, just to grasp a vague graphical understanding of all this? $\endgroup$
    – Eugenio T.
    Jan 19 at 2:11
  • $\begingroup$ To specify, I'm not meaning laying on the axis, just regarding the $>$ and $<$ statements $\endgroup$
    – Eugenio T.
    Jan 19 at 2:13

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