Does $\sum_{n \geq 1}f_n(x)$ converges uniformly on $[0, +\infty)$? Let $f_n(x) = \frac{x^2}{1 + nx^2}\sin(\frac{nx}{x + 1})$.
Is it true that $$ \sum_{n \geq 1} \frac{x^2}{1 + nx^2}\sin(\frac{nx}{x + 1})$$
converges uniformly on $[0, +\infty)$
I want to say 'Yes' and use Abel-Dirichlet test with:
$$a_n(x) = \sin(\frac{nx}{x + 1})$$ and $$b_n(x) = \frac{x^2}{1 + nx^2}$$ which converges to zero. But I have some troubles with proving than $a_n(x)$ and $b_n(x)$ satisfy the conditions we need.
Am I right in my attempts and what is the best way to prove that $f_n(x)$ converges uniformly in this case.

The series converges uniformly on $[0,\infty)$ by the Dirichlet test.
Factoring the terms of  the series as
$$\sum_{n=1}^\infty \underbrace{\frac{x^2}{1+nx^2}}_{b_n(x)} \underbrace{\sin \frac{nx}{x+1}}_{a_n(x)}$$
is not helpful for applying the Dirichlet test.  The requirement that $b_n(x) \to 0$ as $n \to \infty$ both monotonically and uniformly for all $x \in [0,\infty)$ is met.  However, the partial sums $\sum_{n=1}^N a_n(x)$ are not uniformly bounded for all $N \in \mathbb{N}$ and $x \in [0,\infty)$.   This is a consequence of the fact that $\sum_{n=1}^N \sin nz$ is not uniformly bounded for all $z$ on any set  with $2m\pi, \, (m=0,1,2,\ldots)$ as a limit point.
We can apply the Dirichlet test sucessfully by factoring the terms with
$$a_n(x) = \frac{x}{x+1}\sin \frac{nx}{x+1},\quad b_n(x) = \frac{x(x+1)}{1 + nx^2}$$
We have for all $x \in (0,\infty)$,
$$\left|\sum_{n=1}^N a_n(x)\right| = \left|\frac{x}{x+1}\sum_{n=1}^N \sin \frac{nx}{x+1}\right|= \frac{x}{x+1}\frac{\left|\sin\frac{Nx}{2(x+1)}\right|\, \left| \sin \frac{N+1)x}{2(x+1)}\right|}{\left|\sin \frac{x}{2(x+1)}\right|} \\ \leqslant \frac{\frac{x}{x+1}}{\sin \frac{x}{2(x+1)}} \leqslant \frac{1}{\sin 1/2}$$
The last inequality is a consequence of monotonicity and
$$\lim_{x \to 0}\frac{\frac{x}{x+1}}{\sin \frac{x}{2(x+1)}}= 2, \quad \lim_{x \to \infty }\frac{\frac{x}{x+1}}{\sin \frac{x}{2(x+1)}}= \frac{1}{\sin 1/2} \approx 2.0858$$
Hence, the partial sums $\sum_{n=1}^N a_n(x)$ are uniformly bounded.
We also have $\lim_{n \to \infty} b_n(x) = 0$ both montonically and uniformly for all $x \in [0,\infty)$. Clearly $b_n(x)$ decreases with $n$ for fixed $x$.  For uniform convergence note that for all $x \in [0,1)$ we have
$$0 \leqslant b_n(x) = \frac{x(x+1)}{1 + nx^2}= (x+1) \frac{\sqrt{n}x}{1+ nx^2}\frac{1}{\sqrt{n}}\leqslant \frac{2}{\sqrt{n}},$$
and for all $x \in [1,\infty)$ we have
$$0 \leqslant b_n(x) = \frac{x(x+1)}{1 + nx^2}\leqslant \frac{2x^2}{1+nx^2} \leqslant \frac{2}{n}$$
