Evaluate the uniform and the normal convergence of the series of functions $\sum \frac{x}{(1 + x^2)^n}$ 
A series of functions with the general term $ f_n(x) $ defined as:
$$ f_n(x) = \frac{x}{(1 + x^2)^n} \space , \space  x \geq 0  $$

*

*Evalute the uniform and normal convergence of the series $\sum_{n = 1}^{+ \infty} \frac{x}{(1 + x^2)^n} $ on $\mathbb{R}^+$

*Prove that the series have normal convergence on $[a, + \infty[$ with $ a > 0 $

*Is the series uniformally convergent on $[0,1]$?



We have for all $ x \geq 0 $:
$$ f_n'(x) = \frac{(1 + x)^{n - 1}[1 + (1 - 2n)x^2]}{(1 + x^2)^{2n}} = 0 \iff x = \frac{1}{\sqrt{2n - 1}} $$
and for all $ x \geq 0 $ :
$$ | f_n(x) | \leq f_n( \frac{1}{\sqrt{2n - 1}} ) = \frac{\sqrt{2n - 1}}{2^n (2n - 1) (1 + n)^n} \leq \frac{1}{2^n} $$
Since $ \sum \frac{1}{2^n} $ is convergent, then $ \sum f_n $ have normal convergence on $[0, + \infty[$.


*From this question, I think that $ \sum f_n $ have normal convergent on the interval $[a, + \infty[$ only, and not on $ [0, + \infty[ $


*From what I found in question 1, the series have normal convergence on $ [0, + \infty[ $, which implies uniform convergence on the interval $ [0,1] $
What am I missing? Thank you.
 A: For $x=0$, the sum converges to $\sum_{n=0}^\infty f_n(0)=0$.
For $x>0$, we have
$\sum_{n=0}^\infty f_n(x)$ = $x \sum_{n=0}^\infty \left(\frac{1}{1+x^2}\right)^n$
where the expression in parentheses is strictly smaller than 1 for $x>0$, so we have a geometric sum
which equals $x \frac{1}{1-\frac{1}{1+x^2}}=\frac{1+x^2}{x}=:F(x)$ with
$F(x)\rightarrow \infty$ as $x\rightarrow 0^+$
So your function series does not uniformly converge on $[0,a]$ for any $a>0$ but it may be (and is by your criteria) on $[a,\infty)$ for all $a>0$.
(EDIT: above derivation is for sum starting at $n=0$, but same argument for sum starting at $n=1$)
EDIT 2: the above product of $x$ with a geometric series illustrates that formally, for $x=0$, we have a form "$0\times \infty$", so we know, we have to be careful at $x=0$
A: The series is not uniformly convergent for $x \in[0,1]$. A typical approach to showing this (without knowing the sum in closed form) is
$$\tag{*}\sup_{x \in [0,1]}\left|\sum_{k=n+1}^\infty \frac{x}{(1+x^2)^k}\right| = \sup_{x \in [0,1]}\sum_{k=n+1}^\infty \frac{x}{(1+x^2)^k}\geqslant \sup_{x \in [0,1]}\sum_{k=n+1}^{2n} \frac{x}{(1+x^2)^k}\\ \geqslant\sup_{x \in [0,1]}n \cdot \frac{x}{(1+x^2)^{2n}}\geqslant n \cdot \frac{\frac{1}{\sqrt{n}}}{\left(1 + \left(\frac{1}{\sqrt{n}}\right)^2\right)^{2n}}= \frac{\sqrt{n}}{\left(1+ \frac{1}{n}\right)^{2n}}$$
Since $\left(1+ \frac{1}{n}\right)^{2n} \to e^2$ as $n \to \infty$, the RHS of (*) tends to $+\infty$, and, thus,
$$\lim_{n \to \infty}\sup_{x \in [0,1]}\left|\sum_{k=n+1}^\infty \frac{x}{(1+x^2)^k}\right|\neq 0$$
A: The series $\sum_{n = 1}^{+ \infty} \frac{x}{(1 + x^2)^n}$ is a geometric series.
$$
\sum_{n = 1}^{+ \infty} \frac{x}{(1 + x^2)^n} =
x \sum_{n = 1}^{+ \infty} \left(\frac{1}{1 + x^2}\right)^n
$$
Geometric series have a remainder that is explicit. You can actually compute the sum of the series for all $x>0$. Moreover convergence of the series $\sum y^n$, $ y= 1/(1+x^2)$, is normal on any interval $y 
\in [0, a]$ with $a<1$, as is seen by bounding $y$ by $a$.
So the series $\sum_{n = 1}^{+ \infty} \frac{x}{(1 + x^2)^n}$ is normally convergent on all intervals $[a,\infty)$ for $a>0$.
EDIT:
But it is not convergent on $[0,\infty)$.
To show that, we need to find the maximal value of $x \mapsto {x(1+x^2)^{-n}}$ on this interval.
I let you find this maximum by differentiating this function. It is attained in $x= {1\over \sqrt{2n-1}}$. The value of the maximum is
${1\over \sqrt{2n-1}} {1\over (1+ {1\over 2n-1})^n}$.
$$ 
\sup\{x(1+x^2)^{-n} \mid x\in [0,\infty[\} = {1\over \sqrt{2n-1}} {1\over (1+ {1\over 2n-1})^n} \sim C/\sqrt{n}
$$
for some constant $C$. This is the general term of a divergent series.
$$
\sum \left\|x(1+x^2)^{-n}\right\|_\infty = +\infty.
$$
