Math Analysis - Sequence of functions - computing sum of the series 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. 
 A: $$
\sum_{n=0}^N (1-x)x^n = \sum_{n=0}^N x^N - \sum_{n=0}^N (-x) x^n = \sum_{n=0}^N x^N - \sum_{n=0}^N x^{n+1}.
$$
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.
A: 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
