# Proof that a sequence is divergent

Is my proof valid?

Suppose $a_n=\frac{1}{n^2}-\sqrt{n}$. I will now proof that $a_n$ diverges:

$\forall \epsilon > o \exists n^* \forall n \ge n^*: |a_n - L| < \epsilon$

So $|\frac{1}{n^2}|-|\sqrt{n}|-|L|<|\frac{1}{n^2}-\sqrt{n}-L|<\epsilon$

$\frac{1}{n^2}-\sqrt{n}-|L|<\epsilon$

$\frac{1}{n^2}-\sqrt{n}<\epsilon + |L|$

Suppose $z=\sqrt{n}$

$\frac{1}{z^4}-z=\frac{1-z^5}{z^4}<\epsilon+|L|$

$-\frac{z^5}{z^4}<\frac{1-z^5}{z^4}<\epsilon+|L|$

$-z<\epsilon+|L|$

$z>-(\epsilon+|L|)$

$n>(-(\epsilon+|L|))^2=(\epsilon+|L|)^2$

$\epsilon<\sqrt{n}-|L|$ but this is not valid for every $\epsilon$, so contraction and $a_n$ can't be converging for that reason and must be diverging.

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No, it’s not valid: no matter what $\epsilon$ and $L$ are, the inequality $\epsilon < \sqrt{n}-\vert L\vert$ is true for all sufficiently large $n$. Going back up the argument, you can’t possibly get a contradiction from $$\frac1{n^2}-\sqrt{n} < \epsilon + \vert L\vert:$$ the left-hand side is never positive for $n\ge 1$, while the right-hand side is always positive, so the inequality is true whenever $n\ge 1$.

Think about why the sequence diverges: $\dfrac1{n^2}$ approaches $0$, and $\sqrt{n}$ increases without bound, so $a_n$ must tend to $-\infty$. In other words, for any $L$ there must be some positive integer $n_L$ such that $a_n < L$ whenever $n \ge n_L$. The trick is to prove this rigorously.

Your idea of using a proof by contradiction is fine, but you should say what you’re doing. In your argument your third line should have been something like this: ‘Assume to get a contradiction that the sequence converges’. In the argument that I’m suggesting here, the first step is:

Suppose that there is some real number $L$ such that $a_n \ge L$ for all $n$, i.e., such that $\dfrac1{n^2}-\sqrt{n} \ge L$ for all $n$.

What does that imply that might be useful? We know pretty much what $\dfrac1{n^2}$ is doing, so let’s put it on the same side of the inequality as the fixed quantity $L$:

Then $\sqrt{n} \le \dfrac1{n^2}-L$ for all $n$.

Now $\dfrac1{n^2}-L \le 1+\vert L\vert$ (why?), so what happens when $n > (1+\vert L\vert)^2$?

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Thanks! So $1/n^2 \le 1$ for $n>0$. $-L$ is obviously smaller or equal to $|L|$. So I get your last point. But, in the beginning we said 'for all n'. I guess this is in contraction with $n \le (1+|L|)^2$? (because L was fixed). Or am I thinking wrong here? – Kevin Sep 25 '11 at 20:46
@Kevin: We assumed that we had a fixed $L$ such that $a_n \ge L$ for all $n$. This implied that $\sqrt{n}\le\frac1{n^2}-L$ for all $n$, and that gave us our contradiction, because we can show that when $n$ is large enough, $\sqrt{n}$ isn’t $\le\frac1{n^2}-L$. – Brian M. Scott Sep 25 '11 at 20:59