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 $\sinh(nx)/\sinh(n)$...
Uniform convergence involving hyperbolic function
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algebra-precalculus
uniform-convergence
elementary-functions
1 Answers
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$$\frac{\sinh(nx)}{\sinh(n)} = \frac{e^{nx}-e^{-nx}}{e^{n}-e^{-n}} = \frac{e^n}{e^{nx}}\cdot\frac{e^{2nx}-1}{e^{2n}-1}\leq e^{-(n-1)x} $$ since $x\mapsto e^{2nx}$ is an increasing function on $[0,1]$.