Hint: For $m, find a reasonably good estimate for $\dfrac{a_n}{a_m}$. Use this estimate to show that our sequence is Cauchy.
Details: For fun we give an elementary approach to the estimates. Note that if $k \ge 2$, then $1+\frac{1}{k^2}<\frac{1}{1-\frac{1}{k^2}}=\frac{k^2}{(k-1)(k+1)}.$ Let $2\le m. Then $\prod_{k=m}^n \left(1+\frac{1}{k^2}\right)< \prod_{k=m}^n \frac{k^2}{(k-1)(k+1)}.$ Write down the product of the first few terms on the right-hand side, and observe the beautiful cancellations. A little examination shows that $\prod_{k=m}^n \frac{k^2}{(k-1)(k+1)}=\frac{mn}{(m-1)(n+1)}<1+\frac{1}{m-1}.\qquad (\ast)$
Now we complete the proof of convergence. From the inequality in the statement of the problem, we conclude using $(\ast)$ that $0<\frac{a_n}{a_m}<1+\frac{1}{m-1}, \qquad\text{and therefore}\qquad |a_n-a_m| <\frac{a_m}{m-1}. $ But from $(\ast)$ we can see that $a_m<2a_2$. So by picking $m$ large enough, and $n>m$, we can make $|a_n-a_m|<\epsilon$ for any preassigned $\epsilon$, that is, the sequence $(a_n)$ is Cauchy.