is this or (when) does this equality hold for weighted power series $s_n=\sum _{k=0}^n a_k$ for every $n$ and $x\in(0,1)$. 
Let $p=(p_n)$ is a sequence of nonnegative numbers with $p_0>0$ s.t $P_n= \sum_{k=0}^{n}p_k \rightarrow \infty$ as $n \rightarrow \infty$ and $p(x)=\sum _{k=0}^\infty p_kx^k.$
Define $f(x)=\frac{1}{p(x)}\sum _{k=0}^\infty p_ks_kx^k$ exists for each $x\in(0,1)$. 
Is this equality holds or when does it hold?
$$\frac{1}{p(x)}\sum _{k=0}^\infty p_ks_kx^k=\sum _{k=0}^\infty p_ka_kx^k   \ \ \ \ \ (*)$$
For example, if $p_n=1$ for all $n$ then $p(x)=\frac{1}{1-x}$ and it is clear that (*) holds.

The identity $(\ast)$ holds if and only if $\sum\limits_{i=0}^kp_ip_{k-i}a_i=p_k\sum\limits_{i=0}^ka_i$ for every $k\geqslant0$. Thus, $(\ast)$ holds for every sequence $(a_k)$ if and only if $p_k=p_ip_{k-i}$ for every $k\geqslant i\geqslant0$. Since $p_0\ne0$, this yields $p_0=1$ and $p_n=p_1^n$ for every $n\geqslant1$, that is, $p(x)=\frac1{1-\lambda x}$. The condition that $P_n\to\infty$ implies that $\lambda\geqslant1$. Conversely, every such function $p(x)$ solves $(\ast)$.
