Uniform convergence of $\sum_{n=1}^\infty x^{n-1}(1-x)^{2}$

Question

The given series of function is as follow

$$\sum_{n=1}^\infty x^{n-1}(1-x)^{2}$$ prove that given series is uniformaly convergent on $[0,1]$

The solution i tried-The given series form an $G.P$ with ratio $x \leq1$

i.e

$$(1-x)^2+x(1-x)^2+x^2(1-x)^2+...$$

Now if i form partial sum of $n$ terms it will be $$s_n=(1-x)^2 \frac{1-x^n}{1-x}$$ $$\lim_{n\to \infty}s_n=\frac{(1-x)^2}{1-x}$$

after that we get $$s=(1-x)$$

now what can i say about convergence

Because we know that if the series of partial sum is uniform convergence then series is uniform convergence ,but here $s$ is something polynomial type .

Please Help

Answer 1

Option:

Weierstrass M test .

$f_n (x)=x^{n-1}(1-x)^2$;

$f_n'(x)=$

$(n-1)x^{n-2}(1-x)^2-2x^{n-1}(1-x)=0;$

$(n-1)(1-x)-2x=0$;

$x(n+1)=n-1$;

$x=\frac{n-1}{n+1}$;

This is the maximum of $f_n$ on $[0,1]$.(Why?)

(Recall : A continuous function on a compact interval attains its maximum, $f_n(x) \ge 0$ on $[0,1]$)

$f_n(\frac{n-1}{n+1})=$

$(\frac{(1-1/n)^{n-1}}{(1+1/n)^{n-1}})(\frac{2}{n+1})^2 <4/n^2.$

Weierstrass M test : $\sum f_n(x)$ is uniformly convergent on $[0,1]$.

https://en.m.wikipedia.org/wiki/Weierstrass_M-test

Answer 2

The answer above is correct. Other approach would be to show that, since $f$ is continuous and $f(1)=0$ then given $\varepsilon >0$ there exists $\delta >0$ such that for all $x \in (1-\delta, 1],\forall n \in \mathbb{N}: |f(x)|<\varepsilon $.

Now, since $f$ converges uniformly to zero in $[0,1)$ so does in $[0,1-\frac{\delta}{2}]$.

Combining both assertions you get uniform convergence of $f$ in $[0,1]$.