Multiplying summation with same indices and limits What would be 
$(1-\sum \limits_{k=0}^m x^k )(1-\sum \limits_{k=0}^m y^k ) ?$
I dont understand how can I multiply summation of same indices. I checked "multiplication of finite sum (inner product space)" this post but it is different than my case.
Any suggestion?
 A: The index variables $k$ are so-called bound variables. This means that their scope (i.e. range of validity) is determined by their sigma-operator $\sum$ and the operator precedence rules.

The following representations are valid
\begin{align*}
  \left(1-\sum_{k=0}^mx^k\right)\left(1-\sum_{k=0}^my^k\right)&=
  \left(1-\color{green}{\left(\sum_{k=0}^mx^k\right)}\right)\left(1-\color{blue}{\left(\sum_{k=0}^my^k\right)}\right)\tag{1}\\
 &=\left(1-\sum_{k=0}^mx^k\right)\left(1-\sum_{\color{blue}{j=0}}^my^{\color{blue}{j}}\right)\tag{2}
  \end{align*}

Comment:

*

*In (1) we present the scope of each of the index variables somewhat more clearly by using inner parenthesis and the colors green and blue.


*In (2) we denote the index variable of the right-most sum with $j$.

Hint: It is often convenient to give different index variables different names, even if they have no overlapping scope. This usually enhances readability.

A: Let's examine series multiplication first:
$$A=\sum_{i\geq0}a_i=a_0+a_1+a_2+\dots$$
$$B=\sum_{i\geq0}b_i=b_0+b_1+b_2+\dots$$
$$AB=\big(a_0+a_1+\dots\big)B$$
$$AB=a_0B+\big(a_1+a_2+\dots\big)B$$
$$AB=a_0B+a_1B+\big(a_2+a_3+\dots\big)B$$
$$AB=a_0B+a_1B+a_2B+\big(a_3+a_4+\dots\big)B$$
This pattern continues:
$$AB=a_0B+a_1B+a_2B+a_3B+\dots$$
$$AB=\sum_{i\geq0}a_iB$$
Now note the following:
$$a_iB=a_i\sum_{k\geq0}b_k$$
$$a_iB=a_i\big(b_0+b_1+b_2+\dots\big)$$
$$a_iB=a_ib_0+a_ib_1+a_ib_2+\dots$$
$$a_iB=\sum_{k\geq0}a_ib_k$$
Plugging in:
$$AB=\sum_{i\geq0}\sum_{k\geq0}a_ib_k$$
Since $i$ is independent of $k$, and they belong to the same set (namely $\{x\in\Bbb Z:x\geq0\})$, we know that
$$
\begin{align}
AB & = a_0b_0+a_0b_1+a_0b_2+\dots \\
 & + a_1b_0+a_1b_1+a_1b_2+\dots \\ 
 & + a_2b_0+a_2b_1+a_2b_2+\dots \\
 & +\dots 
\end{align}
$$
Which can be greatly abbreviated:
$$AB=\sum_{i,k\in S}a_ib_k$$
Where
$$S=\{x\in\Bbb Z:x\geq0\}=\{0,1,2,\dots\}$$

Now we can move onto something more related to your problem:
$$A=\sum_{i=0}^{m}a_i$$
$$B=\sum_{i=0}^{m}b_i$$
These are just like the case above:
$$AB=\sum_{i,k\in S}a_ib_k$$
Where $$S=\{x\in\Bbb Z:0\leq x\leq m\}$$
And of course the fact
$$(1-A)(1-B)=AB-A-B+1$$
still holds when $A$ and $B$ are series. But one should note:
$$-A=-a_0-a_1-a_2-\dots$$
and not $$-A=-a_0+a_1+a_2+\dots$$
