I am having some difficulty with a real analysis problem.
Suppose $f:[0,\infty)\to[0,\infty)$ is a continuous bijection and consider the series $$\sum_{n=1}^{\infty}\frac{nf(x^2)}{1+n^3(f(x^2)^2)}$$
a) Show the series converges pointwise for all $x\in R$.
b) Show that the series converges uniformly on $[\epsilon,\infty)$ when $\epsilon>0$.
c) Show that the series does not converge uniformly on $R$.
What I have done so far:
a) Let $f_n(x)= \frac{nf(x^2)}{1+n^3(f(x^2)^2)}=\frac{1}{\frac{1}{nf(x^2)^2}+n^2}$. So $\lim_{n\to \infty}f_n(x)=\frac{1}{0+\infty}=0$. Hence i it converges pointwise.
b) For this part I am very lost, I asked for a classmate's help and this is what he told me. So I am not sure if it is correct and if it is, I will have questions.
If $f(0)=0 \implies f_n(0)=0$. Hence$$f(x)=\sum_{n=1}^{\infty}\frac{nf(x^2)}{1+n^3(f(x^2)^2)}\implies f(0)=0$$ Now take $M_n=\frac{n}{1+n^2}=\frac{1}{n+\frac{1}{n^2}}\implies \sum M_n= \sum \frac{1}{n+\frac{1}{n^2}}$. Let $T_n=\frac{n}{n^3}=\frac{1}{n^2}$. Then we see $\lim_{n\to \infty}\frac{M_n}{T_n}=\lim_{n\to\infty}\frac{n^3}{1+n^3}=1\neq0$. By the limit comparison test $\sum M_n$ and $\sum T_n$ both converge or both diverge. Since $\sum T_n=\sum \frac{1}{n^2}$ converges then $M_n$ converges too. Hence, by the Weierstrass M Test $\sum f_n(x)$ converges uniformly on $[\epsilon, \infty)$.
c) Since f is not defined on $(-\infty,0)$ f cannot converge uniformly on all of R. I doubt this is sufficient though.
(a) For pointwise convergence of the series, if $x$ is fixed then
$$\frac{nf(x^2)}{1+n^3(f(x^2))^2} \begin{cases}= 0, & \text{if}\,\,f(x^2) = 0\\ \leqslant \frac{1}{f(x^2)}\frac{1}{n^2}, & \text{if} \,\,f(x^2) \neq 0 \end{cases} $$
Hence, the series converges by the comparison test since $\sum_{n \geqslant 1} n^{-2}$ converges.
(c) Note that
$$\left|S_{2n}(x) - S_n(x)\right| = \sum_{k=n+1}^{2n}\frac{kf(x^2)}{1+k^3(f(x^2))^2} \geqslant n \cdot \frac{nf(x^2)}{1 + 8n^3 (f(x^2))^2}$$
Since $f$ is a bijection onto $[0,\infty)$ there exists a point $x_n \in [0,\infty)$ such that $f(x_n^2) = 1/n$ and
$$\left|S_{2n}(x_n) - S_n(x_n)\right| \geqslant \frac{n^2 \frac{1}{n}}{1 + 8n^3 \frac{1}{n^2}} = \frac{1}{8 + \frac{1}{n}}$$
Since the RHS is converges to $1/8$ as $ n \to \infty$, there exists $N \in \mathbb{N}$ such that for all $n > N$ we have
$$\left|S_{2n}(x_n) - S_n(x_n)\right| > \frac{1}{16}$$
and, therefore, the uniform Cauchy criterion is violated and the convergence cannot be uniform on $\mathbb{R}$ or any subset with $0$ as a limit point.
The uniform Cauchy criterion states that convergence of the series $\sum_{n \geqslant 1} f_n(x)$ is uniform for $x \in D$ if given the partial sum $S_n(x) = \sum_{1\leqslant k \leqslant n} f_k(x) $ the following condition is met. For any $\epsilon > 0$ there exists $N \in \mathbf{N}$ such that for all $m > n > N$ and for all $x \in D$ we have $|S_m(x) - S_n(x)| < \epsilon$.
Try to go back and complete part (b).