On the sequence of function $f_n(x)=n^2x(1-x^2)^n $

Question

Is the sequence of functions $f_n(x)=n^2x(1-x^2)^n $ uniformly convergent over $[0,1]$ ? Is the series $\sum_{n=1}^{\infty} f_n(x)$ convergent with $0<x<1$ ? Is the series uniformly convergent with $0<x<1$ ?

Answer 1

For any fixed $0<x<1$, the series is convergent by the Ratio Test. However, the series does not converge uniformly. Its value $S(x)$ satisfies $S(x)\geq x\sum (1-x^2)^n=x/x^2=1/x$, which is unbounded on $(0,1)$. So, for any partial sum $S_N(x)$, there will always be some $0<x<1$ with $S(x)\geq1/x>S_N(x)+1$. So the convergence is not uniform.

Answer 2

$f_{n}(x)\leq n^{2}(1-x^{2})^{n}$ on $I$ so the pointwise limit is $0$.

$f'_{n}(x)=n^{2}\left [ (1-x^{2})^{n}-2nx^{2}(1-x^{2})^{n-1} \right ]$ and so $x=\frac{1}{\sqrt{2n+1}}$ is the maximum on $I$. At this point, $\overline x_{n}$ we have

$f_{n}(\overline x_{n})=n^{2}\left ( \frac{1}{\sqrt {2n+1}} \right )\left ( \frac{2n}{2n+1} \right )^{n}\to \infty$ as $n\to \infty $ so $f_{n}$ does not converge uniformly on $I$.

Now, if $x=0$ or $x=1$ the series obviously converges to $0$. On the other hand, if $x\in (0,1)$ and $c=1-x^{2}$ then $f_n(x)=n^2x(1-x^2)^n\leq \frac{n^{2}}{c^{n}}$ so the series converges pointwise to an $F\in \mathbb R^{I}$.

The convergence is not uniform on $(0,1)$however:

We have that

$F_{N}(x)=x\sum_{n=0}^{N}n^{2}(1-x^{2})^{n}>x\sum_{n=0}^{N}(1-x^{2})^{n}$ and since $\sum_{n=0}^{\infty }(1-x^{2})^{n}=\frac{1}{x^{2}}$ we see that $$\tag1F(x)>\frac{1}{x}$$

Now fix $N$, let $x_{max}$ be the value of $x\in (0,1)$ such that $F_{N}(x_{max})>F_{N}(x), \forall x\in (0,1)$. Then appealing to $(1)$ we may choose $x\in (0,1)$ such that

$F(x)-F_{N}(x_{max})>1$

which implies that

$F(x)-F_{N}(x)>1$

So we have found, for each $N$, an $x$ for which $F(x)-F_{N}(x)>1$ and so $\left \{ F_{N} \right \}_{N}$ does not converge uniformly.