Determine whether the sequence is convergent or divergent Here is the sequence:

$$a_n = \frac{\sqrt{n^4+1}}{n^2+1}$$

How to do the limit?
Hint. Use equivalents, $\sqrt{n^4+1}\sim n^2$ and $n^2+1\sim n^2$.
Observe that
$$
\lim_{n\to\infty} a_n^2=\lim \frac{n^4+1}{n^4+2n^2+1}=1.
$$
Since $a_n&gt;0$ and the function $\sqrt{x}$ is continuous for $x&gt;0$, it follows that $a_n$ converges to $1$ as well.

Furthermore, the series $\sum_{n=1}^{\infty} a_n$ is divergent (but this might not have been what you were asking for?)