Is this the right way to multiply series?
The following might be useful.
We have \begin{align*} \left(\sum_{m=0}^M a_m\right)\left( \sum_{n=0}^N b_n\right) &=\sum_{m=0}^M a_m\left(\sum_{n=0}^N b_n\right)\\ &=\sum_{m=0}^M\sum_{n=0}^N a_mb_n \end{align*}
Some variants of a special case:
\begin{align*} \left(\sum_{n=0}^N a_n\right)^2&=\left( \sum_{m=0}^N a_m\right)\left( \sum_{n=0}^N a_n\right)\\ &=\sum_{m=0}^N\sum_{n=0}^N a_m a_n\\ &=\sum_{0\leq m,n\leq N} a_ma_n\\ &=\sum_{0\leq m< n\leq N} a_ma_n+\sum_{0\leq n< m\leq N} a_ma_n+\sum_{0\leq m=n\leq N} a_ma_n\\ &=2\sum_{0\leq m<n\leq N}a_ma_n +\sum_{n=0}^N a_n^2\\ &=2\sum_{m=0}^{N-1}\sum_{n=m+1}^{N}a_ma_n +\sum_{n=0}^N a_n^2\\ &=2\sum_{n=1}^N\sum_{m=0}^{n-1}a_ma_n +\sum_{n=0}^N a_n^2\\ \end{align*}
No, it is not true. Intuitively, we can think about multiplication of two finite sums as the sum of all the elements of the matrix $$ \left(\begin{array}{cccc}a_1b_1 & a_1b_2 & \cdots & a_1b_n \\a_2b_1 & a_2b_2 & \cdots & \vdots \\\vdots & \vdots & \ddots & \vdots \\a_nb_1 & \cdots & \cdots & a_nb_n\end{array}\right). $$ But the sum $$ \sum_{k=1}^na_kb_k $$ is the sum of only the elements that are on the diagonal of the matrix.
That's a "specific kind" of product: it is the dot product of the two sequences (or vectors) $(a_1,a_2,\cdots)$ and $(b_1,b_2,\cdots)$.
The "product" of two finite series is $\sum {a_k(\sum b_j)}=\sum{\sum{a_k b_j}}$ (which is the distributive law).