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Using the Weierstrass M-test I can show that the function series $$\sum_{n=1}^{\infty}\frac{n^x}{3^n-5}$$ is uniformly convergent on any closed bounded interval $[0,a]$.

I have a feeling that the series is not uniformly convergent for $x \in [0, \infty)$ since the numerator can not be bounded but how can I show that formally?

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If $$f(x)= \sum_{n=1}^{\infty}\frac{n^x}{3^n-5}$$ was uniformly convergent on $[0,\infty)$, you would have $$\lim\limits_{N \to \infty} \sup_{x \in [0,\infty)} \left\vert R_N(x) \right\vert = 0$$ where $$R_N(x)= \sum_{n=N}^{\infty}\frac{n^x}{3^n-5}$$ But the series is having positive terms for $n \ge 2$. So $$R_N(x) \ge \frac{N^x}{3^N-5} \ge 0$$ And we get a contradiction taking $x_N = N\frac{\ln 3}{\ln N}$ for which $$R_N(x_N) = \frac{3^N}{3^N-5} > 1$$

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Hint: If $\sum f_n$ converges uniformly on a set $E,$ then $f_n \to 0$ uniformly on $E.$

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