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I would like to get a hint on how to establish the convergence of the following sequence:

$$a_{n+1}= a_n + \frac{\sqrt{\vert a_n \vert }}{n^2}$$

where $a_1$ is arbitrary. This is an increasing sequence, so if I could show that it was bounded above I would be done. I cannot figure out how to do that. Any help would be appreciated.

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We have $a_n=\sum_{j=1}^{n-1}\frac{\sqrt{|a_j|}}{j^2}+a_1$. Let $M$ such that $M>|a_1|+\frac{\pi^2}6\sqrt M$. Assume that $|a_j|\leqslant M$ for all $1\leqslant j\leqslant n-1$. Then $$|a_n|\leqslant \sqrt M\sum_{j=1}^{n-1}\frac 1{j^2}+|a_1|<M.$$ As $|a_1|<M$, we are done.

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If $a_1\geq 1$, you have that $\sqrt{a_n}<a_n$ for all $n$. Hence, your sequence $(a_n)$ is dominated by the sequence $(b_n)$, where $$ b_{n+1}=b_n\left(1+\frac{1}{n^2}\right),\quad b_1=a_1. $$

Can you show that $(b_n)$ is bounded?

If $a_1<1$ then either the sequence $(a_n)$ is bounded by $1$, or there exists an index $N$ such that $a_N>1$, and the same argument applies as before.

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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.

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