Proving for all polynomials $p(x)=p_0+p_1x+...+p_dx^d$, $ \ \sum^\infty_{n=1}p(a_n)$ converges iff $p_0=0$ 
The series  $\displaystyle\sum^\infty_{n=1}a_n$ converges absolutely.
Prove that for every polynomial $p(x)=p_0+p_1x+...+p_dx^d$, $\
\displaystyle\sum^\infty_{n=1}p(a_n)$ converges iff $p_0=0$.

What I did: 
$\displaystyle\sum^\infty_{n=1}p(a_n)=\displaystyle\sum^\infty_{n=1}(p_0+...p_dx^d)=\sum^\infty_{n=1}p_0+\sum^\infty_{n=1}\sum^d_{i=1}p_i(a_n)^i$
Let's take a look at $\displaystyle\sum^\infty_{n=1}\sum^d_{i=1}p_i(a_n)^i$, we know $a_n$ converges, raising it to an exponent won't change it's convergence and scaler multiplication won't either, so all 'elements' in the sum except for $p_0$ converge. 
$\displaystyle\sum^\infty_{n=1}p_0$ will always diverge unless $p_0 = 0$.

You're essentially right. The only thing to be careful of is dealing with $\sum_{n=1} ^\infty \sum_{k=1} ^d p_k a_n ^j$. The point is
\begin{align}
\sum_{n=1} ^\infty \sum_{k=1} ^d p_k a_n ^j &= \sum_{k=1} ^d p_k \sum_{n=1} ^\infty a_n ^j \\
&\leq \sum_{k=1} ^d p_k \sum_{n=1} ^\infty |a_n|^k \\
&\ll \sum_{k=1} ^d p_kS_k \\
&\leq d \max_{1 \leq k \leq d} p_k S_k < \infty
\end{align}
where $S_k$ is finite for all $k$. Using the absolute converge is crucial here as, say,  $\sum_{n=1} ^\infty (-1)^n n^{-1/2}$ converges but, obviously, $\sum_{n=1} ^\infty n^{-1}$ diverges.
This fact puts all the burden of convergence on the sum over the $p_0$ and, as you noted, that is infinite if and only if $p_0 = 0$.
