Although not as neat as Davide Giraudo's answer, you may directly apply some convergence test to establish the result.
By redefining $a_1\leftarrow|a_1|$, we may assume that every $a_n$ is nonnegative. Convergence is trivial when $a_1=0$ or $a_n\le1$ for every $n$. So, assume that $a_n>1$ when $n$ is large. Let $b_1=a_1$ and $b_{n+1} = a_{n+1}-a_n$. We want to show that the series $\sum_n b_n$ converges. Now for sufficiently large $n$ and sufficiently small $c>0$, we have
\begin{align}
\frac{b_{n+1}}{b_n} &= \frac{a_{n+1}-a_n}{a_n-a_{n-1}}\\
&=\frac{(n-1)^2}{n^2}\sqrt{\frac{a_n}{a_{n-1}}}\\
&=\frac{(n-1)^2}{n^2}\sqrt{1+\frac{1}{\sqrt{a_{n-1}}(n-1)^2}}\tag{1}\\
&<\frac{(n-1)^2}{n^2}\sqrt{1+\frac{1}{(n-1)^2}}\\
&=\frac{(n-1)}{n^2}\sqrt{(n-1)^2+1}\\
&<\frac{(n-1)}{n^2}(n-1+c).
\end{align}
Therefore $n\left(\frac{b_{n+1}}{b_n}-1\right)<-(2-c)+\frac{1+c}{n}<-1$. As equality (1) implies that $\frac{b_{n+1}}{b_n}\rightarrow1$ when $n\rightarrow\infty$, by Raabe's test, $\sum_n b_n$ converges.