Convergence and uniform convergence?

Question

Let (n=1 -> infinity) x(1-x)^n.

a) Show that this series is convergent on [0,1] to f, where f(0) = 0 and f(x) = 1 for 0<x<=1.

By using the ratio test,...
x(1-x)^n+1/x(1-x)^n = 1-x

So, i find that |1-x|<1, so x is convergent on 0 to 2, so it is obviously convergent on [0,1].
What I don't understand is how to find the proper function where f(0) = 0 and f(x) = 1 on 0<x<=1.

b) Show that this series is not uniformly convergent on [0,1].

I know that there is a theorem for sequences that states that if f isn't continuous, then the sequence isn't uniformly convergent. Does the same hold here?

ecapS trebliH · Accepted Answer

The series is a *geometric series* in powers of y = (1-x),  namely x Sum_(0,oo) y^n,
which converges absolutely to f(x) = x/(1 - y) if |y| < 1 .  (Standard property of 
geometric series. Review:  the ratio test is based on comparison with a geometric series;
it relies on the theory of geometric series. Using the ratio test to check convergence
of a geometric series amounts to comparing the series with itself.)

Note that the sum f(x) =  x/(1-y) = x/x = 1.

Also |y| < 1 means |1-x| < 1  or  0 < x < 2 

If x = 0, all terms of the series are 0. Therefore the series converges to f(0) = 0.

Thus the series converges on [0,2) to f(x), which is not continuous at 0.
Likewise the series converges to f on the closed bounded interval [0,1]

By definition, convergence of a series means convergence of its partial sums,
which are just polynomials in this case; so they are certainly continuous.  Since the limit
f(x) is not continuous, the partial sums do not converge uniformly on [0,1].
This means, by definition, that the series does not converge uniformly on [0,1].

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Convergence and uniform convergence?

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