# Relevance of upper vs lower sum indices

I read this related thread but it doesn't give me a satisfactory answer to the following question:

Must it be true that the order of indices in a sum is relevant? A finite sum is essentially adding up all elements in a set. Who cares it that set is ordered forwards or backwards? Consider the following examples.

Common convention tells us that

$$\sum_{n=5}^0n=0$$

even though we are essentially trying to add the elements of {$n\in[0,5]$} backwards. The validity of this becomes paramount in the following case,

$$\sum_{m=0}^n f(n-m)\ne0$$

when we make a simplifying change of variable $m=n-i$. Then, the sum becomes

$$\sum_{i=n}^0 f(i)=0, \text{by convention.}$$

Obviously nothing significant has changed with the sum when we just change a variable definition, Why, then, does the convention dictate that the value of the sum must change? It seems this particular sum convention is nonsense and should be abandoned.

• Since $i$ appears with a minus sign, the limits should be switched (like in integrals). – egreg Sep 11 '18 at 13:43
• @egreg What do you mean $i$ appears with a sign? $m=n-i$ means $m=0$ maps to $n=i$ as the lower index. – avikarto Sep 11 '18 at 13:47
• You have $n\color{red}{-}i$. Such a transformation reverses the order. – egreg Sep 11 '18 at 13:58

## 1 Answer

If you write the summation as $$\sum_{0\le m\le n}f(n-m)$$ and do the substitution $m=n-i$, then the condition in the sum becomes $$0\le n-i\le n$$ which is equivalent to $$n\ge i\ge 0$$ which in turn can be written as $0\le i\le n$.

Therefore $$\sum_{0\le m\le n}f(n-m)=\sum_{0\le i\le n}f(i)$$

A summation such as $$\sum_{n=5}^0 f(n)$$ can (should?) be written as $$\sum_{5\le n\le0}f(n)$$ and no index $n$ satisfies the condition, so the summation is by definition $0$.

• Thanks, the sum notation in hindsight makes much more sense when written as such an inequality. – avikarto Sep 11 '18 at 14:07
• @avikarto Yes, that's why I always prefer using inequalities in my notes. – egreg Sep 11 '18 at 14:11