Adding Two Power Series if their bounds are different I have the following product
$$\frac{1}{6}\sum_{n=0}^{\infty}\left(\frac{-x}{2}\right)^n\sum_{n=0}^{\infty}\frac{a_nx^{n+2}}{n!}$$
Where $a_n$ is an arbitrary coefficient.  If I factor out $x^2$ from the second power series i can rewrite as a Cauchy Product
$$\frac{x^2}{6}\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{(-1)^ka_{n-k}}{2^k(n-k)!}x^n$$
Now this becomes
$$\frac{1}{6}\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{(-1)^ka_{n-k}}{2^k(n-k)!}x^{n+2}$$
But I'm confused how to reindex the convolution
Is the reindexed summation
$$\frac{1}{6}\sum_{n=2}^{\infty}\sum_{k=2}^{n}\frac{(-1)^ka_{n-k}}{2^{k-2}(n-k)!}x^n$$
EDIT:
There is a bigger picture to this problem.  I have the following that I want to combine as a single sum somehow (pardon the pun):
$$\frac{1}{1+\frac{x}{2}} \sum_{n=0}^{\infty}\frac{a_nx^n}{n!}-\frac{1}{6(1+\frac{x}{2})} \sum_{n=0}^{\infty}\frac{b_nx^{n+2}}{n!}   $$
I know that I can rewrite the first fraction $\frac{1}{1+\frac{x}{2}}$ as $\frac{1}{1-\frac{-x}{2}}$ which can be rewritten as a power series 
$$\frac{1}{1-\frac{-x}{2}}=\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{2^n}$$
Rewriting the difference and taking Cauchy Product of both terms yields
$$\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{2^n}\sum_{n=0}^{\infty}\frac{a_nx^n}{n!}-\frac{1}{6}\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{2^n}\sum_{n=0}^{\infty}\frac{b_nx^{n+2}}{n!}   $$
$$\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{(-1)^ka_{n-k}}{2^k(n-k)!}x^n-\frac{1}{6}\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{(-1)^kb_{n-k}}{2^k(n-k)!}x^{n+2}$$
$$\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{(-1)^ka_{n-k}}{2^k(n-k)!}x^n-\frac{1}{6}\sum_{n=2}^{\infty}\sum_{k=2}^{n}\frac{(-1)^kb_{n-k}}{2^{k-2}(n-k)!}x^n$$
How do I procede from there?
 A: Your reindexed summation is correct. Since the sums (outer and inner) for the second series start at $n = 2$ resp. $k = 2$ while in the first they start at $0$, combining the two power series is not totally direct. One way is to split the sums, treat $n = 0$ and $n = 1$ separately, and in the remaining series, for the inner sum $k = 0$ and $k = 1$. A more compact notation can be obtained using Iverson brackets,
$$[P] = \begin{cases} 1 &, P\text{ is true}\\ 0 &, \text{otherwise}.\end{cases}$$
Then we can write the second series
$$\frac{1}{6}\sum_{n=2}^\infty \Biggl(\sum_{k=2}^n\frac{(-1)^k b_{n-k}}{2^{k-2}(n-k)!}\Biggr)x^n = \sum_{n=0}^\infty [n\geqslant 2]\Biggl(\sum_{k=0}^n \frac{[k\geqslant 2]2(-1)^kb_{n-k}}{2\cdot 2^k(n-k)!}\Biggr)x^n$$
and combine as
$$\sum_{n=0}^\infty\Biggl(\sum_{k=0}^n \frac{(-1)^k}{2^k(n-k)!}\biggl(a_{n-k} - \frac{2}{3}[n\geqslant 2][k\geqslant 2] b_{n-k}\biggr)\Biggr)x^n.$$
Since we have $k\leqslant n$, the condition $n \geqslant 2$ is implied by $k\geqslant 2$, hence redundant, and we can further simplify to
$$\sum_{n=0}^\infty\Biggl(\sum_{k=0}^n \frac{(-1)^k}{2^k(n-k)!}\biggl(a_{n-k} - \frac{2}{3}[k\geqslant 2] b_{n-k}\biggr)\Biggr)x^n.$$
