the power series converges in compact convergence topology

Question

Consider the sequence of functions $f_{n}: (-1,1) \rightarrow R$ defined by:$$f_{n}(x) = \sum_{k=1}^{n}{kx^{k}}$$
a) Prove that $(f_{n})$ converges in the topology of compact convergence, conclude that the limit function is continuous
b)Show that $(f_{n})$ does not converge in the uniform topology.

I'm learning function space in topology and find it hard to imagine, especially the compact convergence topology. I got stuck in this problem

As I know, we don't even have the clear formula for the function that this sequence converges to, so how can we prove that sequence converges at all. One more thing, is there other ways to prove a sequence of functions converges in compact convergence topology except proving it converges uniformly in every compact subspace, cause in some spaces, we don't know all their compact subspaces

Thanks.

Answer 1

In this case you can actually determine what the limit has to be, if it exists. The series $\sum_{k\ge 1}kx^k$ is absolutely convergent for $|x|<1$. We know that $$\sum_{k\ge 0}x^k=\frac1{1-x}$$ for $|x|<1$; taking the derivative and multiplying by $x$ yields $$\sum_{k\ge 1}kx^k=\frac{x}{(1-x)^2}\;,$$ which is the only reasonable candidate for the limit of the sequence $\langle f_n:n\in\Bbb Z^+\rangle$.

Note that

$$\sum_{k=1}^nkx^k=\sum_{k\ge 1}kx^k-\sum_{k\ge n+1}kx^k\;,$$

and you can use the same trick to get a closed form for $\sum_{k\ge n+1}kx^k$, so you can actually get an explicit formula for each $f_n$.

Finally, every compact subset of $(-1,1)$ is contained in a closed interval $[-a,a]$ with $0<a<1$, so you need only look at these intervals.

Answer 2

$$F(x)=\sum_{k=0}^n x^k=\frac{1-x^n}{1-x}\iff F'(x)=\sum_{k=0}^n kx^{k-1}=\sum_{k=0}^n kx^k-\sum_{k=0}^n x^k=\sum_{k=0}^n kx^k-F(x)$$ $$f(x)=\sum_{k=0}^n kx^k=F(x)+F'(x)=\frac{1-x^n}{1-x}+\left(\frac{1-x^n}{1-x}\right)_x'=\ldots$$