Operation order sum vs addition

Question

As a reader seeing an equation like $$ \sum_{i=1}^3 1+1 $$ (as "everybody" writes it without parenthesis)

Is it

1. Sum before Addition: $\sum_{i=1}^3 1+1=\left(\sum_{i=1}^3 1\right)+1=4$
2. Addition before Sum: $\sum_{i=1}^3 1+1=\sum_{i=1}^3 \left(1+1\right)=6$

Sometimes it is imho used even conversely

• $\sum_{i=1}^3 i+1$ is often assumed to be $\left(\sum_{i=1}^3 i\right)+1=7$
• however $\sum_{i=1}^3 1+i$ can only be calculated as: $\sum_{i=1}^3 \left(1+i\right)=9$

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Update: Let's take an realistic example from https://en.wikipedia.org/wiki/Interatomic_potential#Functional_form

$$ V_\mathrm{TOT} = \sum_{i}^N V_1(\vec r_i) + \sum_{i,j}^N V_2(\vec r_i,\vec r_j) + \sum_{i,j,k}^N V_3(\vec r_i,\vec r_j,\vec r_k) + \cdots $$

is it

1. $V_\mathrm{TOT} = \left[\sum_{i}^N V_1(\vec r_i)\right] + \left[\sum_{i,j}^N V_2(\vec r_i,\vec r_j)\right] + \left[\sum_{i,j,k}^N V_3(\vec r_i,\vec r_j,\vec r_k)\right] + \cdots$
2. $V_\mathrm{TOT} = \sum_{i}^N \left(V_1(\vec r_i) + \sum_{j}^N \left[V_2(\vec r_i,\vec r_j) + \sum_{k}^N \left\{V_3(\vec r_i,\vec r_j,\vec r_k) + \cdots\right\}\right]\right)$

I know the answer: $$V_\mathrm{TOT} = \left[\sum_{i}^N V_1(\vec r_i)\right] + \left[\sum_{i=1}^{N-1} \sum_{j=i+1}^N V_2(\vec r_i,\vec r_j)\right] + \left[\sum_{i}^{N-2} \sum_{j=i+1}^{N-1} \sum_{k=j+1}^N V_3(\vec r_i,\vec r_j,\vec r_k)\right] + \cdots$$

It is very often that someone writes $\sum_{i=1}^3 a_i+\sum_{j=1}^3 b_j$, letting the reader decide if b_1,b_2,b_3 should be counted once or three times.

As a writer I could make parantheses, but as a reader I can't.

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Update2:

• In $\sum_i A_i + \sum_j a_{ij}$ most would assume $\sum_i \left[A_i + \sum_j a_{ij}\right]$
• but in $\sum_i A_i + \sum_j a_{j}$ most would assume $\left[\sum_i A_i\right] + \sum_j a_{j}$

I think equations should not be inperpreted differently based on there use, I think any valid form of equation should be unambiguous.

Answer 1

$$ \sum_{i=1}^3 1+1 $$ Is it

1. Sum before Addition: $\sum_{i=1}^3 1+1=\left(\sum_{i=1}^3 1\right)+1=4$
2. Addition before Sum: $\sum_{i=1}^3 1+1=\sum_{i=1}^3 \left(1+1\right)=6$

It is actually 3. Confusing and unclear.

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There is no established rule about how far the summation operator reaches. The most important reason is that usually, when we are summing something like $$\sum_{i=1}^3 a_i$$ the expression $a_i$ depends on the value of $i$, and if we are careful so as not to overload the variable $i$, it should be clear that basically, the summation operator reaches as far as is needed so that the expression makes sense. For example,

$$\sum_{i=1}^3 i + i$$ doesn't make sense if it is interpreted as $$\left(\sum_{i=1}^3 i\right) + i$$ because the value of $i$ is then left hanging.

I would say most of the time, the sum ends at the first $+$ symbol, so in your case, I would say that, if you force them to decide, most mathematicians will say the result is $4$.

But most mathematicians will likely first say that this is a needlessly unclear way of writing the sum, and will tell you to either use parentheses or reorder the sum to make it clear. For example,

$$1 + \sum_{i=1}^3 1$$ is unambiguously equal to $4$.