How do I prove a sequence diverges to infinity?

Question

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.

Answer 1

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!