# Diverging sequence

I can't understand diverging sequences. How can I prove that $a_n=1/n^2-\sqrt{n}$ is divering? Where to start? What picture should I have in my mind? I tried to use $\exists z \forall n^* \exists n\ge n^*: |a_n-A|\ge z$, but how should I see this? And how can I solve the question with this property?

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Do you mean $a_n = 1/(n^2-\sqrt n)$ or $a_n = (1/n^2) - \sqrt n$? – Asaf Karagila Sep 25 '11 at 18:08
I meant $a_n=(1/n^2)-sqrt(n)$ – Kevin Sep 25 '11 at 18:09
Just making sure. It is best to use $\frac{x}{y}$ to remove ambiguity in such cases. – Asaf Karagila Sep 25 '11 at 18:10

There is no algorithm that accepts an arbitrary sequence $(a_n)_{n\geq1}$ as input and provides a proof of its divergence as output. Finding such a proof is a matter of professional experience: Having seen so many sequences in exercices, having a toolbox of asymptotic estimates at ones disposal, etc. In the example at hand one immediately observes that the given sequence $(a_n)_{n\geq1}$ is unbounded, as $\sqrt{n}$ gets arbitrarily large when $n$ gets large, while ${1\over n^2}$ converges to $0$.

Now we have to convert this idea into a formal proof. We have to show that for any given $M>0$ there is an $n$ such that $$a_n<-M\ .\qquad (*)$$ The condition $n>M^2$ guarantees $-\sqrt{n}<-M$, but we have to take care of the ${1\over n^2}$ in the definition of $a_n$. Now ${1\over n^2}\leq1$, therefore $$a_n={1\over n^2}-\sqrt{n}< 1-(M+1)=-M$$ is true as soon as $n>(M+1)^2$. This means that the condition $(*)$ is true not only for ${\it some}\ n$, but actually for ${\it all}\ n>M+1$. Therefore the sequence $(a_n)_{n\geq1}$ is not only unbounded (whence divergent), but actually "improperly convergent" to $-\infty$.

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I think there should be $\forall A$ as well in your expression as you want to say that no matter what limit somebody proposes, it fails. For your series,as $n$ gets large, $1/n^2$ gets small, but $-\sqrt{n}$ gets large and negative. Given $A, z$ and $n^*$, you need to find an $n$ so that $a_n$ is farther from $A$ than $z$. The largest (in absolute value) that $a_n$ can be and fit is $|A|+z$, so can you find a good $n$?

To prove there is no limit, suppose somebody proposes a limit $A$. We will choose $z=1$ and prove that for $n>n^*, |a_n-A|\gt 1$. Let us choose $n^*=\min((|A|+2)^2,2),$ then for $n \gt n^*, a_n \lt a_{n^*}$, then $|a_{n^*}-A|\gt |\frac{1}{n^{*^2}}-\sqrt{n^*}-A|$ $\gt |A|+2-|A|-1/2 \gt 3/2$ so $A$ is not a limit.

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A convergent sequence is also a Cauchy sequence. If you can find a constant $\epsilon>0$ so that for all $N>0$, there are $n>N$ and $m>N$ so that $|a_n-a_m|\ge\epsilon$, then $\{a_n\}$ is a divergent sequence.

In the case of $a_n=\frac{1}{n^2}-\sqrt{n}$, we can let $\epsilon=1$ and for a given $N>0$, let $n=N+1$ and $m=N+2+\lceil2\sqrt{N+1}\;\rceil$. Then \begin{align} &a_n-a_m\\ &=\left(\frac{1}{(N+1)^2}-\sqrt{N+1}\right)-\left(\frac{1}{(N+2+\lceil2\sqrt{N+1}\;\rceil)^2}-\sqrt{N+2+\lceil2\sqrt{N+1}\;\rceil}\right)\\ &=\left(\frac{1}{(N+1)^2}-\frac{1}{(N+2+\lceil2\sqrt{N+1}\;\rceil)^2}\right)+\left(\sqrt{N+2+\lceil2\sqrt{N+1}\;\rceil}-\sqrt{N+1}\right)\\ &\ge1 \end{align} since $$0\le\frac{1}{(N+1)^2}-\frac{1}{(N+2+\lceil2\sqrt{N+1}\;\rceil)^2}$$ and \begin{align} 1&=\sqrt{N+2+2\sqrt{N+1}}-\sqrt{N+1}\\ &\le\sqrt{N+2+\lceil2\sqrt{N+1}\;\rceil}-\sqrt{N+1} \end{align}

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Now I got this:

$|a_n-A| \ge \epsilon$

$\frac{1}{n^2} - sqrt(n) \ge \epsilon + |A|$

Suppose $u=sqrt(n) (u \ge 0)$

$u^{-4}-u \ge \epsilon + |A|$

$u^{-4} \ge u^{-4} - u$

$u^{-4} \ge \epsilon + |A|$

$u \ge (\epsilon + |A|)^\frac{-1}{4}$

$n \ge \frac{1}{sqrt(\epsilon + |A|)}$

And what may I conclude now?

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Your second line is incorrect. For large $n$, the left side will be negative and cannot be greater than the right. I will put some more into my answer. – Ross Millikan Sep 29 '11 at 18:00