Alternative index notation for summation

Question

I didn't want to continue the discussion in comments so I decided to ask as a question:

What do we gain by writing this simple expression

$\sum_{k=0}^n 2^k = 2^{n+1} - 1$

as this where final index is missing and the index $k$ is a range:

$\sum_{0\le k < n} 2^k = 2^{n} - 1$

Apparently this is used in high mathematics by professional mathematicians but what do we gain in this case?

Answer 1

There are at least two (related) advantages to excluding the upper bound.

1. The sum $$\sum_{m \leq k < n}$$ (with $k$ being the bound variable) extends over $n - m$ terms; if you include the upper bound, it’s $n - m + 1$.

2. If you split up a summation, you get $$\sum_{m \leq k < n} = \sum_{m \leq k < l} + \sum_{l \leq k < n}$$ for any $m \leq l \leq n$. If you include the upper bound, you get shifts by $1$ again.

(For these reasons, it’s common for range operators in programming languages to work this way.)

Now, in math the notation $$\sum_{k=m}^n$$ is firmly established to include the upper bound. If you want to exclude it, you have the option of using either the notation from above; or you need to change the meaning of the summation sign (which will be confusing even if you are explicit about it); or you need to invent your own version, like $$\sum_{k=m}^{<n}.$$

Of course, the notation with $m \leq k < n$ in the subscript also has a technical problem: It is not apparent anymore which variable is the one bound by the summation sign. I have seen $$\sum_{k : m \leq k < n}$$ as a possible notation to indicate that, but often, context is enough.

Answer 2

Of course, $$\begin{align}\forall n\in\{0,1,\dots\}\quad\sum_{k=0}^n 2^k = 2^{n+1} - 1\\\iff\forall n\in\{1,2,\dots\}\quad\sum_{k=0}^{n-1}2^k = 2^n- 1.\end{align}$$ This can be rewritten: $$\begin{align}\forall n\in\{0,1,\dots\}\quad\sum_{0\le k\le n}2^k = 2^{n+1} - 1\\\iff\forall n\in\{1,2,\dots\}\quad\sum_{0\le k\le n-1}2^k = 2^n- 1\\\iff\forall n\in\{1,2,\dots\}\quad\sum_{0\le k<n}2^k = 2^n- 1.\end{align}$$

As explained many times in comments to your previous post which you are refering to, the advantage of $\sum_{0\le k < n} 2^k = 2^n- 1$ over $\sum_{k=0}^{n-1} 2^k = 2^n- 1$ was for $n=0$: $$\sum_{0\le k <0}\text{anything}=\sum_{k\in\varnothing}\text{anything}=0$$ (cf. "Empty sum" on Wikipedia), whereas, as far as I know, $\sum_{k=0}^{-1} 2^k$ makes no sense.

This allowed to prove by induction $$\forall n\in\{\color{red}0,1,2,\dots\}\quad\sum_{0\le k<n}2^k = 2^n- 1,$$ the base case being $0=2^0-1$.