There are different types of convergence when you are talking about functions. You didn't specify the type; the following is a proof for POINTWISE convergence.

pointwise convergence means:

fix x. (with uniform convergence you don't get to do this, and that makes it harder)

fix `epsi`

Now, with both of those fixed, you find an `N` such that, for every `n>N`

`|f_n(x)-0| < epsi `

We will do this case by case. First fix `x=0` and show convergence, then fix `x=1` and show convergence, then fix `0<x<1` and show convergence.

Let `x=0` and fix `epsi`

Then, for every n, `f_n(x)=0` , so we may pick `N` to be any natural number, since every `n` will work.

Thus, for `x=0`, `f_n(x) -> 0`

Now, let `x=1`, and fix `epsi`

Again, for every n, `f_n(x)=0` , so we may pick `N` to be any natural number, since every such `n` will work.

Thus, for `x=1`, `f_n(x) -> 0`

Now, fix `x`, where `0<x<1`, and fix `epsi`

We use a ratio test for sequences.

The theorem we will use is:

`lim_(n->oo) |(a_(n+1))/(a_n)|=L<1 => a_n -> 0`

`f_(n+1) = (n+1)x(1-x^2)^(n+1)`

`f_n(x) = nx(1-x^2)^n `

`(f_(n+1))/(f_n)= (n+1)/(n)(1-x^2)`

The limit of this, as `n -> oo` is `(1-x^2)`

Since `0<x<1` , this number is between 0 and 1

So, using the theorem, we have that, for `0<x<1`, `f_n(x) -> 0`

So, for each `0<=x<=1`, we have that `f_n(x) ->0`

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mlehuzzah | Certified Educator