Uniform convergence involving hyperbolic function

Question

I need to prove that the series $$\sum_{n=1}^{\infty}\frac{\sinh(nx)}{n \sinh(n)}$$ is uniformly convergent in $[0,1-\epsilon]$ for all $0<\epsilon\leq1$. I have tried with the Weierstrass test but I get nowhere since I don't know hot to bound the term $\sinh(nx)/\sinh(n)$...

Answer 1

$$\frac{\sinh(nx)}{\sinh(n)} = \frac{e^{nx}-e^{-nx}}{e^{n}-e^{-n}} = \frac{e^n}{e^{nx}}\cdot\frac{e^{2nx}-1}{e^{2n}-1}\leq e^{-(n-1)x} $$ since $x\mapsto e^{2nx}$ is an increasing function on $[0,1]$.