Now we have $displaystylesum_{n=1}^{infty}$$dfrac{1 }{2n }$x$^{2n}$ the place x$in$R. We let 0≤a<1.

I’ve to point out that the sum perform $f$:[-a,a]–>R is differentiable with $f'(x)=frac{x}{1-x^2}$ for $xin [-a,a]$.

I believe I can use this Theorem:

If we let $displaystylesum_{n=1}^{infty}$$f_n$ be a sequence of steady features on [a,b].

And if we assume that $displaystylesum_{n=1}^{infty}$$f’_n$ is uniformly convergent on [a,b] and there exist a c$in$[a,b] so $displaystylesum_{n=1}^{infty}$$f_n(c)$ is convergent; then $displaystylesum_{n=1}^{infty}$$f_n$ is uniformly convergent on [a,b]

with a differential sum perform

the place it applies that ($displaystylesum_{n=1}^{infty}$$f_n(x)$)’=$displaystylesum_{n=1}^{infty}$$f’_n(x)$.

I’ve discovered that $displaystylesum_{n=1}^{infty}$$frac{partial }{partial x}$$(dfrac{1 }{2n }$x$^{2n})$= $frac{x}{1-x^2}$ for x$in$[-a,a], however how can I present it?

And is it sufficient utilizing this theorem to point out that the sum perform is differential?

And in that case how can I show that $displaystylesum_{n=1}^{infty}$$dfrac{1 }{2n }$x$^{2n}$ is a sequence of steady features on [-a,a]?

Hope anybody can provide me a touch?