# Determine the values of $a$ in the series $\sum_{n=0}^{\infty} x^{a}e^{-nx}$ for which $s(x)$ is continuous on $[0,\infty)$

Consider the series $\sum_{n=0}^{\infty} x^{a}e^{-nx}$ where $a$ is a positive constant.

(a) Prove that the series converges point-wise for all $x\geq0$ and determine its sum $s(x)$.

(b) Determine the values of $a$ for which $s(x)$ is continuous on $[0,\infty)$.

(c) Which conclusions can you draw about uniformity of convergence?

So far I have for part (a) that as $n\rightarrow \infty$, $x^{a}e^{-nx}\rightarrow 0$ which implies point-wise convergence, and that $\sum_{n=0}^{\infty} x^{a}e^{-nx}$ is a geometric series with sum $s(x)=\frac{x^{a}}{1-e^{-x}}.$ For part (c) I understand that if we can show that $s(x)$ is not continuous for particular values of $a$ then it is impossible for it to be uniformly convergent for those values of $a$. Assuming I am correct with the above? I am not sure on how to proceed with part (b)? Any help greatly appreciated.

• "So far I have for part (a) that as $n\rightarrow \infty$, $x^{a}e^{-nx}\rightarrow 0$ which implies point-wise convergence" No, that only shows a test for divergence fails. – zhw. Aug 7 '17 at 3:25
• Agghhh, dont know what I was thinking. I should have applied the Weierstrass M-test.? – Eiraus Aug 7 '17 at 3:33