Suppose $g: \mathbb R_{\ge 0}\to \mathbb R_{\ge 0}$ is a continuous bijective function. Prove that the series $$\sum_{k\ge 1}\frac{kg(x^2)}{1+k^3[g(x^2)]^2}$$ is not uniformly convergent on $\mathbb R$.
If I knew the sum of the series, I could use the negation of the definition of the uniform convergence of partial sums, but since I don't know the limit of partial sums (and how to find one), I cannot do this. How can I proceed?
Let $g$ be a continuous, bijective function from $\mathbb{R}_{\ge0}\to\mathbb{R}_{\ge0}$. Furthermore, let $S(x)$ be the function given by the series
$$S(x)=\sum_{n=1}^\infty \frac{ng(x^2)}{1+n^3g^2(x^2)}$$
where we denote the $N$'th partial sum of $S(x)$ by $S_N(x)$. Clearly, $S(x)$ is convergent for every $x$.
We wish to show that the convergence fails to be uniform on $[0,\infty)$.
Note that $S(x)$ does converge uniformly on $[x_0,\infty)$ for all $x_0>0$.
Inasmuch as $g$ is bijective, we can let $y=g(x^2)$ and analyze the uniform convergence of the series $\sum_{n=1}^\infty \frac{ny}{1+n^3y^2}$ for $y\in [0,\infty)$.
To show that the convergence is not uniform for $y\in [0,\infty)$, we will use the negation of uniform convergence, namely that there exists an $\epsilon>0$ such that for all $N\ge 1$ there exists a number $N'>N$ and a $y\in [0,\infty)$ such that
$$\left|\sum_{n=1}^\infty \frac{ny}{1+n^3y^2}-\sum_{n=1}^N\frac{ny}{1+n^3y^2}\right|\ge \epsilon$$
Proceeding, we have with $y=1/(N+1)^{3/2}$
$$\begin{align} \left|\sum_{n=1}^\infty \frac{ny}{1+n^3y^2}-\sum_{n=1}^{N}\frac{ny}{1+n^3y^2}\right|&=\sum_{n=N+1}^\infty \frac{ny}{1+n^3y^2}\\\\ &=\sum_{n=N+1}^\infty \frac{n/(N+1)^{3/2}}{1+\left(\frac{n}{N+1}\right)^3}\\\\ &\ge \sum_{n=N+1}^{2N+2}\frac{n/(N+1)^{3/2}}{1+\left(\frac{n}{N+1}\right)^3}\\\\ &\ge N\frac{1}{9(N+1)^{1/2}}\\\\ &\ge \frac{1}{9\sqrt 2} \end{align}$$
So, we have found a number $\epsilon=\frac{1}{9\sqrt 2}>0$, such that for all $N$, there exists a number $y=1/(N+1)^{3/2}$ such that
$$\left|\sum_{n=1}^\infty \frac{ny}{1+n^3y^2}-\sum_{n=1}^N\frac{ny}{1+n^3y^2}\right|\ge \epsilon$$
which negates the uniform convergence. And we are done!