Sequence of functions and function series For every $n \ge 0$ we define function $f_n:[-1;1]\rightarrow \mathbb{R}$
$f_n(x) = \sqrt[n+1]{n+1}(\frac{x+x^2}{2})^n$, $x\in[-1;1]$, $0^0=1$.


*

*a)determine whether sequence of functions $\{f_n\}$ converges uniformly

*b)show that function $S:(-1;1)\rightarrow\mathbb{R}$ 


$S(x) = \sum_{n=0}^{+\infty}{f_n(x)}$, $x\in(-1;1)$ is well-defined. Compute $S(0)$ and its derivative (if exists)
$\{f_n\}$ converges pointwise to $0$, but $\lim_{n\to\infty}{sup_{x\in[-1;1]}{\sqrt[n+1]{n+1}(\frac{x+x^2}{2})^n}} = 1$ so sequence cannot converge  uniformly. But how to solve subpoint b)?
 A: Note that for a fixed $x\in(-1,1)$ we have $\frac{|x+x^2|}2 \leq \frac {|x| + |x^2|}{2} < \frac {2|x|}2 < 1$ and combine that with the fact that $n^{\frac 1n}$ converges to 1 to obtain the pointwise convergence of the series using the Weierstrass theorem. Actually it converges uniformly on every interval of the form $(-1+\varepsilon, 1-\varepsilon)$ $(0 < \varepsilon < 1)$ since $\frac {|x + x^2|}2 < 1-\varepsilon$ therein.  The series of derivatives looks as follows $$T(x) = \sum\limits_{n=0}^\infty f_n'(x) = \sum\limits_{n=1}^{\infty} (n+1)^{\frac 1{n+1}} n (\frac{x+x^2}2)^{n-1}(x+\frac 12).$$ A slight modification of the argument for S(x) shows that T(x) is uniformly convergent in $(-1+\varepsilon, 1-\varepsilon)$. Thus $T(x) = S'(x)$ in some neighbourhood of 0. 
A: Denote $$u_n=\sqrt[n+1]{n+1}.$$ $u_n$ is converging to $1$ hence is bounded by a number $A >0$ ($u_n < A$ for all $n \in \mathbb N$).
For $0 < R < 1$, you have for $x \in [-R,R]$: $$\left\vert (\frac{x+x^2}{2})^n \right\vert \le \vert x \vert^n (\frac{1+ \vert x \vert}{2})^n \le \vert x \vert^n \le R^n$$ Hence the series $\sum f_n$ is normally and therefore uniformly convergent on all compacts $[-R,R]$. As $f_n(1)=u_n$, $(f_n)$ is not uniformly convergent on $[-1,+1]$.
As a consequence of above points, $S$ is continuous on $(-1,+1)$ and $S(0)=1$.
Using similar techniques as above we get $$\left\vert \frac{f_n(x) - f_n(0)}{x} \right\vert \le A.\vert x \vert^{n-1}$$ for $x \in [-R,+R]$ and $n \ge 1$. Consequently $$\lim\limits_{x \to 0} \frac{\sum_{n \ge 2} f_n(x) - \sum_{n \ge 2} f_n(x)}{x} = 0$$ Also $$\lim\limits_{x \to 0} \frac{f_1(x) - f_1(0)}{x} = \frac{\sqrt{2}}{2}$$ Finally $S^\prime(0)=\frac{\sqrt{2}}{2}$.
Regarding $S$ differentiability in general, one can prove that $\sum f_n^\prime$ is normally convergent on $(-1,+1)$. As $\sum f_n$ is pointwise convergent on the same interval, $S$ is differentiable on $(-1,+1)$.
