Algebraic rules involving series: scalar multiplication and addition

We now deal with the basic operations of addition and (scalar) multiplication of series.

Suppose we must deal with the series of the addition of two sequences, such as

\[\sum_{r=1}^{n} (u_r + v_r)\]

Unpacking yields:

\[(u_1 + v_1) + (u_2 + v_2) + \dots + (u_n + v_n)\]

Grouping up the \(u_r\)'s (respectively, the \(v_r\)'s) together as one block, we have

\[(u_1 + u_2 + \dots + u_n) + (v_1 + v_2 + \dots + v_n)\]

Thus, it follows that

\[\sum_{r=1}^{n} (u_r + v_r) = \sum_{r=1}^{n} u_r + \sum_{r=1}^{n} v_r\]

Now we deal with the operation of scalar multiplication:

\[\sum_{r=1}^{n} \alpha u_r\]

where \(\alpha\) is a constant (called a scalar), and \(\{u_r\}\) is a given sequence.

Unpacking this sigma notation, we have

\[\alpha u_1 + \alpha u_2 + \dots + \alpha u_n\]

and we can factor out the common factor of \(\alpha\):

\[\alpha (u_1 + u_2 + \dots + u_n) = \alpha \sum_{r=1}^{n} u_r\]

In short, if \(\alpha\) is a constant (scalar) then

\[\sum_{r=1}^{n} \alpha u_r = \alpha \sum_{r=1}^{n} u_r\]

Here, we must emphasize that \(\alpha\) is a constant, and thus in particular does not depend on \(r\).