Prove that $\Sigma_{k = 0}^\infty e^{-kx}$ converges uniformly on any closed subinterval of $(0,\infty)$
attempt: by the Weierstrass M-Test if $I$ is a nonempty subset of $ R$ and let $f_k: I → R$ for all natural $k$. Suppose that $M_k \geq 0$ satisfies $\Sigma_{k = 0}^\infty M_k< \infty$. If $|f_k(x)| \leq M_k$ for natural $k$, and $x \in I$, then $\Sigma_{k = 0}^\infty f_k$ converges a absolutely and uniformly on I.
Then let $[a,b]$ be a subinterval of $(0,\infty)$, and let $f_k = e^{-kx} = (\frac{1}{e^k})^x$. can anyone please help me choose $M_k$? I don't know how to continue. Thank you in advance.
Let $[a,b]$ be a closed subinterval of $(0,\infty)$. Then on $[a,b]$, $|e^{-kx}| = e^{-kx} \le e^{-ka}$. Since $\sum_{k = 1}^\infty e^{-ka}$ converges, the Weierstrass $M$-test ensures $\sum_{k = 0}^\infty e^{-kx}$ converges uniformly on $[a,b]$.