<s>[INST] 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?
 [/INST] 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
$$
\begin{align}
1&=\sqrt{N+2+2\sqrt{N+1}}-\sqrt{N+1}\\
&\le\sqrt{N+2+\lceil2\sqrt{N+1}\;\rceil}-\sqrt{N+1}
\end{align}
$$
</s>