I have searched Approach Zero for a similar question, but without success. I am trying to verify the uniform convergence of $$\sum^\infty_{k=0}\frac{x^k}{1+x^k},$$ on $-\frac12\leq x\leq \frac12$. I get stuck pretty early on in my approach and it makes me wonder if it's the way to go. I want to use the Weierstrass $M$-test and so my aim is to find an upper bound of the absolute value of the terms in the series. Put $f_k(x)=x^k/(1+x^k)$, then $$f_k '(x)=\begin{cases}0 \qquad \text{if } k=0 \\ \frac{kx^{k-1}}{(1+x^k)^2}\quad \text{if } k\geq 1.\end{cases}$$
This derivative confuses me. I'm unsure how to find the critical points in $(-1/2,1/2)$. I appreciate any help and feel free to suggest a different approach.
Note that for $k \geq 1$ and $x \in [-1/2,1/2]$, $$ \left|\frac{x^k}{1 + x^k} \right| \leq \frac{|x|^k}{1 - |x|^k} \leq \frac{|x|^k}{1 - \frac 12} = 2|x|^k \leq 2^{1-k}. $$ Similarly, assuming that we define $x^0 = 1$ for all $x$, we have $\frac{x^0}{1 + x^0} = \frac 12 \leq 2^{1 - 0}$. Thus, you can apply the Weierstrass $M$-test to $f_n(x) = \frac{x^k}{1 + x^k}$, taking $M_k = 2^{1-k}$.