Showing that $\frac{1+nx^2}{(1+x^2)^n}$ converges to $0$ I aim to show that $\{a_n\} = \frac{1+nx^2}{(1+x^2)^n}$ converges to $0$.  The following two facts seem obvious:
(1) $\forall n \in \mathbb{N}, a_n \ge 0$ (since each $a_n$ is the product of two positive numbers).
(2) $\forall n \in \mathbb{N}, a_n \ge a_{n+1}$.
Yet this is not quite enough to show that $a_n \rightarrow 0$ since the sequence could very well converge to some other $c > 0$.

The problem is false for $x=0$. If $x \neq 0$, then
$$\lim_n \frac{1+nx^2}{(1+x^2)^n} = \lim_n \frac{1}{(1+x^2)^n}+x^2\lim_n \frac{n}{(1+x^2)^n}$$
It is trivial to show that
$$\lim_n \frac{1}{(1+x^2)^n} =0$$
While, for $x\neq 0$ we have
$$\lim_n \frac{n}{(1+x^2)^n} = \lim_n \left( \frac{\sqrt[n]{n}}{(1+x^2)} \right)^n=0$$
the last equality following from the fact that
$$ \lim_n  \frac{\sqrt[n]{n}}{(1+x^2)} = \frac{1}{x^2+1} <1$$
