Let $f(x)$ be the sum function of the power series
$$\sum_{n=0}^{\infty} a_nx^n$$
on $(-R,R)$ for $R>0$. If $f(x)+f(-x)=0$
for all $x∈(-R,R)$, prove that $a_n=0$ for all even $n$
Now What I have done is as follows:-
As we know, $1/(1+x) =$ $$\sum_{n=0}^{\infty} (-1)^nx^n$$
Similarly I write,
$$\sum_{n=0}^{\infty} a_nx^n$$ $=1/(1-a_nx^n)=f(x)$, $x∈(-R,R)$
then $f(-x)=$ $$\sum_{n=0}^{\infty} (-1)^na_nx^n$$
$=1/(1+a_nx^n)$, $x∈(-R,R)$
On the basis of given information we have
$f(x)+f(-x)=0$
$→$ $$\sum_{n=0}^{\infty} a_nx^n$$ $+$ $$\sum_{n=0}^{\infty} (-1)^na_nx^n$$ $=0$
$→$ $1/(1-a_nx^n)+1/(1+a_nx^n)=0$
$→$ $2/(a_nx^n)=0$
$→$ $2*1/a_n*x^-n=0$
My question is from here can we conclude that the $a_n=0$ for all even $n$ ?
And if it is not the correct way please someone suggest me how to proceed. Thank you in advance.
Here is my second approach
If I break the general terms in this way then the following stuffs come:-
If $f(x)= $$\sum_{n=0}^{\infty} a_nx^n$
$→$ $=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+...$
On the other hand I write
$f(-x)= $$\sum_{n=0}^{\infty}(-1)^n a_nx^n$
$→$ $=a_0-a_1x+a_2x^2-a_3x^3+a_4x^4-a_5x^5+...$
Now from $f(x)+f(-x)=0$ we have
$2*[a_0+a_2x^2+a_4x^4+a_6x^6+...]=0$
$→$ $2*$$\sum_{n=0}^{\infty} a_2nx^{2n}=0$
Now $x^{2n}$ are positive terms so that $a_n=0$ for even $n$. Is it correct?