Even Power Series properties Let $$\sum_{k=0}^{\infty}h_kx^k$$ be a series with convergence radius $R>0$ and sum function $s:(-R,R)\rightarrow \mathbb{R}$.

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*Show that if $h_{2k+1}=0$ for alle $k\geq 0$ then $s$ is even and that
$$\sum_{k=0}^{\infty}h_k(1-(-1)^k)x^k$$ has a sum $s(x)-s(-x)$.

Use the uniqueness of a power series to conclude that if $s$ is even, then $h_{2k+1}=0$ for all $n\geq0$.

I know that a function is even if $g(x)=g(-x)$. Furthermore, I knwo that for 2 $\sum_{n=0}^{\infty}a_nx^n,\sum_{n=0}^{\infty}b_nx^n$ that both have convergence radius (atleast) $r>0$ and that $$\sum_{n=0}^{\infty}a_nx^n=\sum_{n=0}^{\infty}b_nx^n$$
holds for all $x\in (-r,r)$ then $a_n=b_n$ for all $n\in \mathbb{N_0}$.
Even though I understand these results I am struggeling solving this problem as it seems vague; i.e not specific.
 A: Assume that $H:(-R,R)\to\mathbb{R}$ is even: for all $x\in(-R,R)$, $H(-x)=H(x)$ \emph{i.e.}
\begin{align*}
H(-x)&=\sum_{n=0}^{+\infty}h_n(-x)^n=\sum_{k=0}^{+\infty}h_{2k}(-x)^{2k}+\sum_{k=0}^{+\infty}h_{2k+1}(-x)^{2k+1}=\sum_{k=0}^{+\infty}h_{2k}x^{2k}-\sum_{k=0}^{+\infty}h_{2k+1}x^{2k+1},\\
H(x)&=\sum_{n=0}^{+\infty}h_nx^n=\sum_{k=0}^{+\infty}h_{2k}x^{2k}+\sum_{k=0}^{+\infty}h_{2k+1}x^{2k+1}.
\end{align*}
This yields after subtracting the second line from the first one:
\begin{align*}
0&=H(-x)-H(x)=-\sum_{k=0}^{+\infty}h_{2k+1}x^{2k+1}-\sum_{k=0}^{+\infty}h_{2k+1}x^{2k+1}=-2\sum_{k=0}^{+\infty}h_{2k+1}x^{2k+1}.
\end{align*}
The sum $\widetilde{H}$ of the series $\sum h_{2k+1}x^{2k+1}$ is thus identically zero over $(-R,R)$. As
\begin{align*}
h_{2k+1}&=\frac{1}{(2k+1)!}\widetilde{H}^{(2k+1)}(0)
\end{align*}
and $\widetilde{H}\equiv0$, we deduce that $h_{2k+1}=0$ for all $k\in\mathbb{N}$.
Assume conversely that $h_{2k+1}=0$ for all $k\in\mathbb{N}$. Then for all $x\in(-R,R)$,
\begin{align*}
H(x)&=\sum_{n=0}^{+\infty}h_nx^n=\sum_{k=0}^{+\infty}h_{2k}x^{2k}+\sum_{k=0}^{+\infty}\underbrace{h_{2k+1}}_{=0}x^{2k+1}=\sum_{k=0}^{+\infty}h_{2k}x^{2k},\\
H(-x)&=\sum_{n=0}^{+\infty}h_n(-x)^n=\sum_{k=0}^{+\infty}h_{2k}(-x)^{2k}+\sum_{k=0}^{+\infty}\underbrace{h_{2k+1}}_{=0}(-x)^{2k+1}=\sum_{k=0}^{+\infty}h_{2k}(-x)^{2k}=\sum_{k=0}^{+\infty}h_{2k}x^{2k}
\end{align*}
i.e. $H(-x)=H(x)$ and $H$ is even.
