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Use the Weierstrass's test to determine a set on which the series converges absolutely uniformly $$\sum nx^n(1-x)^n$$

The test is as follow:

The series $\sum f_n$ converges uniformly on S if $$ ||f_n||_S \leq M_n.$$ With $ n\geq k $, and where $\sum M_n < \infty$

(a) Pointwise:

when $x\in (0,1)$, $\lim\limits_{n \rightarrow \infty} (1-x)^n =0$ and $\lim\limits_{n \rightarrow \infty} x^n =0 \implies \lim\limits_{n \rightarrow \infty} ( nx^n(1-x)^n) =0 $

When $x=\{0,1\} \implies \lim\limits_{n \rightarrow \infty} ( nx^n(1-x)^n) =0 $

When $x<0$, $\lim\limits_{n \rightarrow \infty} (1-x)^n = \infty $ and $\lim\limits_{n \rightarrow \infty} x^n = +/- \infty \implies \lim\limits_{n \rightarrow \infty} ( nx^n(1-x)^n) = +/- \infty $

When $x>1$, $\lim\limits_{n \rightarrow \infty} (1-x)^n = +/-\infty $ and $\lim\limits_{n \rightarrow \infty} x^n = \infty \implies \lim\limits_{n \rightarrow \infty} ( nx^n(1-x)^n) = +/- \infty $ It follows that it can only converge in the interval $S=[0,1]$ of the limit function f(x) =0

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(b) Let's evaluate the norm:

$$ ||f_n||_S = \sup ( nx^n(1-x)^n) $$

$$\frac{d}{dx} ( nx^n(1-x)^n) = n^2(1-x)^n x^{n-1}- n^2(1-x)^{n-1} x^n = \frac{n^2(1-x)^nx^{n-1}(2x-1)}{x-1}$$

Critical points: $x=0, x=1/2, x=1$,

After testing $x=1/4 \implies \frac{d}{dx} ( nx^n(1-x)^n) >0$,

and testing $x =3/4 \implies \frac{d}{dx} ( nx^n(1-x)^n) <0$

It follows that $x={0,1}$ are local minimum and $x=1/2$ a local maximum

$$ ||f_n||_{S=[0,1]} = \sup ( nx^n(1-x)^n) = n(1/2)^n(1-(1/2))^n= n(1/2)^{2n} $$

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(c) Let's find an upper bound that convergences. As when $x \in [0,1]$: $$x(1-x) < x$$ $$x^n(1-x)^n < x^n$$ $$nx^n(1-x)^n < nx^n$$

(d) $\sum nx^n =\frac{n}{1-x}$ converges in $|x|<1$ and diverges in $|x| \geq 1$. It follows

$$||f_n||_{S=[0,1)} = n(1/2)^{2n} \leq nx^n = M_n$$

Therefore, $\sum nx^n(1-x)^n$ is uniformly convergent in $S=[0,1)$

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Am I going in the right direction? if not what needs to be rectified?

Thx for your input

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  • $\begingroup$ Just as a comment the root test does very well here. $\endgroup$ – user223391 Aug 14 '17 at 13:49
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The maximum of $x(1-x)$ on $[0,1]$ is $1/4$, achieved at $x=1/2$. Therefore, the maximum of $nx^n(1-x)^n$ on $[0,1]$ is $n/4^n$, achieved at $x=1/2$.

As $\sum_{n=1}^\infty n/4^n$ is absolutely convergent, then on $[0,1]$, $\sum_{n=1}^\infty nx^n(1-x)^n$ is absolutely and uniformly convergent.

This argument can be extended to larger intervals. If there are $a<0$, $b>1$ with $-1<\min(a(1-a),b(1-b))$ then the sum will be absolutely and uniformly convergent on $[a,b]$ (again using the M-test).

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