How can I add such two series?

Question

I know that for two series $\space \sum^{L}_{n=0} a_n \space$ and $\space \sum^{L}_{k=0} b_k$ we can say

$$\sum^{L}_{n=0} a_n \space + \sum^{L}_{k=0} b_k= \sum^{L}_{j=0} {a_j} + {b_j}\\$$

But what if we had the following addition?

$$\sum^{M}_{n=1} a_n \sum^{M}_{\rho =0} nb_{\rho} \space + \space \sum^{M}_{n=1} s_n \sum^{M}_{\rho =0} nt_{\rho} \space$$

Where

$$\sum^{M}_{n=1} a_n \sum^{M}_{\rho =0} nb_{\rho}=\sum^{M}_{n=1} \sum^{M}_{\rho =0} a_nnb_{\rho}$$

and

$$\sum^{M}_{n=1} s_n \sum^{M}_{\rho =0} nt_{\rho}=\sum^{M}_{n=1} \sum^{M}_{\rho =0} s_nnt_{\rho}$$

How would I add these type of series seeing as they share similar indexes?

Answer 1

Here are some aspects you could consider:

The sigma symbol $\sum$ is a short hand notation for \begin{align*} \sum_{n=0}^La_n=a_0+a_1+\cdots+a_L \end{align*} The index $n$ is a so-called bounded variable, which means it is bounded to the sigma operator and is not of any use otherwise.

Contrary to this running index $n$ are the lower limit $0$, which is a constant and the upper limit $L$ of the series, which is a free variable not bounded to the sigma operator. They can also be seen as index of the first and last summand on the right hand side.

We are free to change the names of bounded variables as long as there is no naming conflict in the expression with other items. We can write the same series using another index variable. We can write e.g. \begin{align*} \sum_{n=0}^La_n=\sum_{k=0}^La_k&=a_0+a_1+\cdots+a_L\\ &=a_L+a_{L-1}+\cdots+a_0=\sum_{k=0}^{L}a_{L-k} \end{align*}

In the expression above I have also exchanged the order of summation in the last two representations, which is perfectly valid since for any two reals $a,b$ addition is commutative (and associative). \begin{align*} a+b=b+a \end{align*}

The representation in your first line has a little ambiguity. It does not represent what you presumably have in mind. You should write instead \begin{align*} \sum_{n=0}^La_n+\sum_{k=0}^La_k=\sum_{j=0}^L\left(a_j+b_j\right) \end{align*}

Addition is usually considered to be left-associative meaning the operations are grouped from left to right. Without brackets, the right hand side of your first line is then interpreted as \begin{align*} \sum_{j=0}^La_j+b_j=\left(\sum_{j=0}^La_j\right)+b_j=a_0+a_1+\cdots+a_L+b_j \end{align*} We see, the sum is taken in this case only over the $a_j$ and the summand $b_j$ is considered as variable with a free index variable $j$ (and without any meaning in your context).

Hint: Use always brackets in such cases in order to avoid ambiguities.

When finding alternate representations of the second expression

\begin{align*} \sum_{n=1}^Ma_n\sum_{\rho=0}^Mnb_\rho+\sum_{n=1}^Ms_n\sum_{\rho=0}^Mnt_{\rho} \end{align*}

we can also apply the distributive law $a(b+c)=ab+ac$ for real numbers.

Here are some of many different possibilities:

\begin{align*} \sum_{n=1}^M&a_n\sum_{\rho=0}^Mnb_\rho+\sum_{n=1}^Ms_n\sum_{\rho=0}^Mnt_{\rho}\\ &=\sum_{n=1}^Mna_n\sum_{\rho=0}^Mb_\rho+\sum_{n=1}^Mns_n\sum_{\rho=0}^Mt_{\rho}\\ &=\sum_{n=1}^M\left(na_n\sum_{\rho=0}^Mb_\rho+ns_n\sum_{\rho=0}^Mt_{\rho}\right)\\ &=\sum_{n=1}^Mn\left(a_n\sum_{\rho=0}^Mb_\rho+s_n\sum_{\rho=0}^Mt_{\rho}\right)\\ &=\sum_{n=1}^Mn\sum_{\rho=0}^M\left(a_nb_\rho+s_nt_{\rho}\right)\\ &=\sum_{n=1}^M\sum_{\rho=0}^Mn\left(a_nb_\rho+s_nt_{\rho}\right)\\ &=\sum_{{n=1}\atop{\rho=0}}^Mn\left(a_nb_\rho+s_nt_{\rho}\right)\\ \end{align*}

Note: When manipulating finite series you are free to rearrange, reorder the summands as long as you follow the operator rules of commutativity, associativity and distributivity and you correctly respect the algebraic sign of the numbers.

Answer 2

From what I see, this is a possible solution.

$\sum^{M}_{n=1} \sum^{M}_{\rho =0} a_nnb_{\rho}+\sum^{M}_{n=1} \sum^{M}_{\rho =0} s_nnt_{\rho}=\frac{\sum^{M}_{n=1} \sum^{M}_{\rho =0} (a_n-s_n)n(b_{\rho}-t_{\rho})+\sum^{M}_{n=1} \sum^{M}_{\rho =0} (a_n+s_n)n(b_{\rho}+t_{\rho})}{2}$