Let:
$$f(n) = n/(n^4+1)$$
if we factor f to:
$$ n/[n^3(n+1/n^3)] $$
Given that this is equivalent, we perform distribution:
$$1/[n^2(n+(1/n^3)]$$
We now have:
$$ 1/(n^3+1/n) $$
So:
$ 1/n ->$ 0 as $n -> \infty $
As well:
$ \frac 1{n^3} -> 0$ as $n -> \infty $
However, the answer uses arctangent and u substituion to solve. I did see a manipulation of arctan's integrand.
Intuitivly, I thought factoring would be easier to find how the function behaves, e.g., diverges/converges.
i would rewrite the term like following $\frac{n}{n^4}\cdot \frac{1}{1+\frac{1}{n^4}}$ and we have $\frac{1}{n^3}\cdot \frac{1}{1+\frac{1}{n^4}}$ and for $n->\infty$ we obtain $0 \cdot \frac{1}{1+0}=0$
Sonnhard.