I have a problem determining when a series of functions converges uniformly or not.
Let $$ f_n(x) = \frac{nx}{1 + n^3 x^{\alpha}} \ \mbox{ for } \ x, \ \alpha \in (0, \infty) \ \mbox{ and } \ n \in \mathbb{N} $$ I have shown that $ f_n $ converges pointwise to zero for any $ \alpha > 0 $, and uniformly to zero for $ \alpha \in [1, 3) $. So the series $ \sum_n f_n(x) $ can converge uniformly only for that values of $ \alpha $.
For $ \alpha = 1 $, the inequality $$ f_n(x) \leq \frac{nx}{n^3 x^{\alpha}} = \frac{1}{n^2} \cdot x^{1 - \alpha} $$ ensures that $ \sum_n f_n $ converges uniformly, because this sum is less than $ \sum_n 1/n^2 $.
For $ \alpha > 1 $ each $ f_n $ has a maximum at $ x = 1/((\alpha - 1) n^3) $ with value $$ f_n \left( \frac{1}{(\alpha - 1) n^3} \right) = \frac{(\alpha - 1)^{1 - 1 / \alpha}}{\alpha} \cdot \frac{1}{n^{-1 + 3 / \alpha}} $$ If also $ \alpha < 3/2 $, then $ -1 + 3 / \alpha > 1 $ and the associated series converges uniformly: $$ \sum_n f_n(x) \leq \sum_n f_n \left( \frac{1}{(\alpha - 1) n^3} \right) = \frac{(\alpha - 1)^{1 - 1 / \alpha}}{\alpha} \cdot \sum_n \frac{1}{n^{-1 + 3 / \alpha}} < \infty $$ In general, for $ \alpha > 1 $, it converges uniformly at any interval of the form $ [\epsilon, \infty) $ with $ \epsilon > 0 $, $$ f_n(x) \leq \frac{nx}{n^3 x^{\alpha}} = \frac{1}{n^2} \cdot x^{1 - \alpha} \leq \frac{1}{n^2} \cdot \epsilon^{1 - \alpha} \longrightarrow \sum_f f_n(x) \leq \epsilon^{1 - \alpha} \sum_n \frac{1}{n^2} < \infty $$ So, I feel the series $ \sum_n f_n(x) $ does not converges uniformly at $ (0, \infty) $ for $ \alpha \in [3/2, 3) $, and the problem should be at $ x = 0 $. Now I think about bound down the sums $$ \sum_{n \geq N} f_n(x) $$ Because if the series really converges uniformly, that sums should be as small as I want (for $ N $ large and for any $ x $), but I haven't it. Any hints on this?
Let $N$ be a natural number. $$\sum_{n=1}^\infty f_n(x)\ge \sum_{n=N}^{2N} f_n(x)$$ For $x_N=N^{-3/\alpha}$ and $N\le n\le 2N$ we have $$f_n(x_N)\ge {N\,N^{-3/\alpha}\over 1+8N^3N^{-3}}={1\over 9}N^{1-3/\alpha}$$ Therefore $$\sum_{n=N}^{2N} f_n(x_N)\ge{1\over 9}N^{2-3/\alpha}$$ For $\alpha\ge {3\over 2}$ the last expression does not tend to $0.$ Therefore for $\alpha\ge {3\over 2}$ the series is not uniformly convergent.
The proof of uniform convergence for $1\le \alpha<{3\over 2}$ can be made simpler. We have $$f_n(x)={nx\over 1+n^3x^\alpha}={(n^3x^\alpha)^{1/\alpha}\over 1+n^3x^\alpha}\,n^{1-3/\alpha}$$ The function $$t\mapsto {t^{1\over \alpha}\over 1+t}$$ is bounded, say by $M_\alpha.$ Then $$0\le f_n(x)\le M_\alpha n^{1-3/\alpha}$$ Therefore by the Weierstrass test the series is uniformly convergent.