We’re given the next sequence

$$

sum_{n=1}^{infty}frac{1}{2n}*x^{2*n} ;;;;;;; ; xin R

$$

and let $0leq a<1$

Present the sequence convergence uniformly within the interval [-a,a]

$$sum_{n=1}^{infty}frac{x^{2n}}{2n}$$

$$f(x)=frac{x^{2}}{2}+frac{x^{4}}{4}+frac{x^{6}}{6}+frac{x^{8}}{8} +….$$

$$f(x)=frac{1}{2}[frac{(x^{2})}{1}+frac{(x^{2})^{2}}{2}+frac{(x^{2})^{3}}{3}+frac{(x^{2})^{4}}{4} +….]$$

$$ln(1-x)=-[x+frac{x^{2}}{2}+frac{x^{3}}{3}+……]$$

$$f(x)=frac{-1}{2}ln(1-x^{2})$$

$$f(x)=frac{1}{2}ln(1-x^{2})$$

My downside is that I received this assist from an individual from Chegg and I do not perceive if it is appropriate or not. And why it’ll make sense .

I’d clear up it by making a

Later on this query I’m requested:

b) present the sequence sum operate f:(-a,a)$rightarrowR$ is differentiabel and it applies

$$f'(x)=frac{x}{1-x^{2}} ;;;;;;;xin(-a,a) $$

$$(f(x))’=(frac{1}{2}*ln(1-x^{2}))’$$

the constante

$$f'(x) = frac{1}{2}*frac{1}{1-x^{2}}^{2x}$$

I substitute $u=1-x^2$ and $f = ln(u)$.

$$frac{1}{2}*frac{1}{u}(-2x)$$

$$f'(x)=frac{x}{1-x^{2}}$$

c) Decide the sumfunctionen f, and use this to point out

$$sum_{n=1}^{infty}frac{1}{2^{n}*n} =log(2)$$

I exploit at fraction rule to re-write

$$sum_{n=1}^{infty}frac{1}{2^{n}*n} $$

til

$$sum_{n=1}^{infty}frac{frac{1}{2}^{n}}{n}$$

$$frac{frac{1}{2}^{1}}{1}+frac{frac{1}{2}^{2}}{2}+….$$

$$ln(1-x)=-[frac{x}{1}+frac{x^{2}}{2}+frac{x^{3}}{3}+….]$$

$$ -ln(1-frac{1}{2})$$

$$-ln(frac{1}{2})=$$

$$=ln(2)$$