Uniform Convergence

1.
(1) x in [0,1), ∑(n=0~∞) (1-x)x^n= (1-x)/(1-x)=1
(2) if x=1, ∑(n=0~∞) (1-x)x^n= 0
so, f(x)=∑(n=0~∞) (1-x)x^n is not conti. in [0,1]
while (1-x)x^n is conti. in [0,1],
hence ∑(n=0~∞) (1-x)x^n is not conv. on [0,1].

2.
Let f(x)= x/(1+n^4 x^2), then max f(x) = f( 1/n^2 )= 1/(2n^2).
Thus, ∑(n=1~∞) x/(1+n^4 x^2) < ∑(n=1~∞) 1/(2n^2) conv. uniformly (By M-test)

3.
(1) For any x in (-1, 1), there exists N>0 such that -1+ 1/N < x < 1- 1/N.
For n> 2N(indep. on x) we have |(x+ 1/n)|^n < (1- 0.5/N)^n,
∑(n=2N~∞) (1- 0.5/N)^n converges ( |ratio|<1),
thus, ∑(n=1~∞) (x+ 1/n)^n conv. uniformly for any x in [-1+1/N,1+1/N].
Hence, ∑(n=1~∞) (x+ 1/n)^n conv. and conti. in (-1,1).

(2) Suppose ∑(n=1~∞) (x+ 1/n)^n converges to f(x) uniformly, then
for any x in (-1,1), there exists N>0 such that ∑(n=N~∞) (x+ 1/n)^n < 1.
But (1+ 1/n)^n -> e= 2.718..., so that we can choose x sufficiently close
to 1, and lim(n->∞) (x+ 1/n)^n > 2
contraticts to ∑(n=N~∞) (x+ 1/n)^n < 1.
Hence, ∑(n=1~∞) (x+ 1/n)^n is not convergent uniformly on (-1,1).