Define a sequence of functions $\{f_n\}$ on $[0,1]$ by
$$ f_n(x) = \left\{ \begin{array}{ll} \frac{1}{n} & \quad \frac{1}{2^{n+1}}< x \leq \frac{1}{2^n} \\ 0 & \quad \text{elsewhere} \end{array} \right. $$
Prove that $\sum_{n=1}^\infty f_n(x)$ converges uniformly on $[0,1]$, but that the Weirstrass M-Test fails.
My attempt:
I've noticed that $$ \sum_{n=1}^\infty f_n(x) = \left\{ \begin{array}{ll} \frac{1}{1} & \quad \frac{1}{2^{2}}< x < \frac{1}{2} \\ \frac{1}{2} & \quad \frac{1}{2^{3}}< x < \frac{1}{2^2}\\ \frac {1}{3} & \quad \frac{1}{2^{4 }}< x < \frac{1}{2^3}\\ \end{array} \right. $$
so I said $|S_n(x) - S_m(x)| = |\sum_{k={m+1}}^n f_k(x)| = \frac{1}{k} $ for some k with $m+ 1 \leq k \leq n$
Then $|\sum_{k={m+1}}^n f_k(x) = \frac{1}{k} < \frac{1}{m+1} < \epsilon$ if $m+ 1 < \frac{1}{\epsilon}$ or $ m > \frac{1}{\epsilon} -1$
Therefore for $\epsilon > 0, \ \ $ choose m to be a positive integer $> \frac{1}{\epsilon} -1$ then it satisties the cauchy criterion.
Does anyone have a better way to do this proof or can fix any errors I have made.