Math Analysis - Sequence of functions - computing sum of the series

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Consider the sequence of functions $f_n:[0,1]$, $f_n:=(-1)^n(1-x)x^n$, $n=0,1,2,\ldots$, $0\le x\le1.$

a. Find the sum $\sum_{n=0}^\infty |f_n(x)|$

b. Show that the series $\sum_{n=0}^\infty |f_n(x)|$ does not converge uniformly on $[0,1]$.

I'm looking for any help/suggestions. Thank you.

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$$ \sum_{n=0}^N (1-x)x^n = \sum_{n=0}^N x^N - \sum_{n=0}^N (-x) x^n = \sum_{n=0}^N x^N - \sum_{n=0}^N x^{n+1}. $$ The last sum above $\displaystyle\sum_{n=0}^N x^{n+1}$, is $$ x^{0+1} + x^{1+x} + x^{2+1} + x^{3+1} + \cdots + x^{N+1} = \sum_{n=1}^{N+1} x^n. $$ So we have $$ \sum_{n=0}^N (\cdots\cdots) - \sum_{n=1}^{N+1} (\cdots\cdots) $$ $$ = \Big(n=0\text{ term}\Big) + \sum_{n=1}^N (\cdots\cdots) - \sum_{n=1}^N (\cdots\cdots) - \Big(n=N+1\text{ term}\Big) $$ $$ = \Big(n=0\text{ term}\Big) - \Big(n=N+1\text{ term}\Big) = x - x^{N+1}. $$ Now you want $\lim\limits_{N\to\infty}$ of that last expression.

When $x=1$, the limit is $0$; when $x < 1$, the limit is $x$. The discontinuity can be used to show that the convergence is not uniform.

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Well for (a) $(1-x)$ is just a constant, let's pull it out, we just have $(1-x)\sum\limits_{n=0}^{\infty} x^n$, well what is that?

note if it converges uniformly, then its limit is the same as the pointwise limit. what does it converge to pointwise? so take a suitable $\epsilon$, no matter how big you make $N$, there will be some $x$ such that it is more than $\epsilon$ away from the pointwise limit