Is this the right way to multiply series? Is 
$$\sum a_n \sum b_n = \sum a_nb_n$$
for finite sums?
In particular, is 
$$\sum a_n \sum a_n = \sum a_n^2?$$
Thanks,
 A: 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).
A: Don't take it badly, but a trivial counter-example is enough:
$$(1+1)(1+1)\ne 1^2+1^2.$$
A: To avoid confusion, it helps to never use the same summation index for two distinct sums. The summation index is a bound variable: it is meaningless outside of the summation, and can be swapped out with any other (unused) variable and still be the same mathematical expression. 
Instead of $$\sum a_n\sum b_n$$ you should write $$\sum_n a_n \sum_m b_m$$ (it also helps to denote the summation variable on the summation sign). 
Now, if you combine the two, you get $$\sum a_n b_m.$$ Here you see there are two summation indices, so you have to have two sums, i.e., $$\sum_{n,m}a_n b_m = \sum_n \sum_m a_n b_m.$$ 
If you use the same summation variable for two distinct summations you can arrive at incorrect conclusions, as you have done above (the same holds for integration variables of definite integrals, by the way). 
A: 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*}

A: 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.
A: Given $\mathrm a := (a_1, a_2, \dots, a_m)$ and $\mathrm b := (b_1, b_2, \dots, b_n)$,
$$\left(\sum_{i=1}^m a_i\right) \left(\sum_{j=1}^n b_j\right) = (1_m^T \mathrm a) (1_n^T \mathrm b) = (1_m^T \mathrm a) (\mathrm b^T 1_n) = 1_m^T (\mathrm a \mathrm b^T) 1_n = \mbox{tr} (1_n 1_m^T \mathrm a \mathrm b^T) = \langle 1_m 1_n^T , \mathrm a \mathrm b^T \rangle$$
Thus, the product of the sums is the sum of the $m n$ entries of $m \times n$ matrix $\mathrm a \mathrm b^T$.
