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?
 A: 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.
A: 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 $-\infty$.
A: 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
$$
\begin{align}
1&=\sqrt{N+2+2\sqrt{N+1}}-\sqrt{N+1}\\
&\le\sqrt{N+2+\lceil2\sqrt{N+1}\;\rceil}-\sqrt{N+1}
\end{align}
$$
A: 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?
