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How does one evaluate something like the following? $\sum_{k=0}^{-1}\left( 5\times 2^k \right)$

When I type this into Mathematica it returns 0. Can someone explain why this is?

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    Seems like the wikipedia article needs some rewriting.2011-04-25

3 Answers 3

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The sum $\sum_{k=a}^b f(k)$ for integers $a$ and $b$ is usually (often, but not exclusively) interpreted as the sum of all values of $f(k)$ for $k$ an integer, $a\leq k\leq b$.

If $a\gt b$, then there are no integers $k$ that satisfy the condition $a\leq k\leq b$, so this is a sum with no summands (or "empty sum").

In order to make sure that associativity laws are respected, empty sums (sums with no summands) are defined to be equal to $0$ (likewise, empty products are defined to be equal to $1$).

Since your sum is empty, it is equal to zero by definition.

See also: this question (discussing the value of a "product with no factors"); the same argument for why the product with no factors "should" be equal to $1$ applies to see why the sum with no summands "should" be equal to $0$, based on associativity.

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    @user9325: Then perhaps you can clarify that sentence, given that there is no "the comment" to refer to any more, while you spend some time explaining what Mathematica does when the upper limit is smaller than the lower limit (saying it treats it as an explicit sum as you do does not answer the *first* question asked by the OP, which is how to evaluate a sum of terms where the first term has larger argument than the last term). You know, geese and gander and all that. Or don't. I *do* begin to understand, though, some of your comments in meta.2011-04-26
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Mathematica treats a sum $\sum_{k=a}^{b}f(k)$ as the sum $f(a)+f(a+1)+\cdots + f(a+\lfloor{b-a}\rfloor)$ which is one possible interpretation that allows sums with non-integer arguments. You can write Sum[f[k],{k,0,-1,-1}] to get the result of the comment.

Another common interpretation is that this sum runs over all integers between $a$ and $b$.

Another common and useful interpretation is the one that for integers $a,b,c$ respects $\sum_{k=a}^{b-1}f(k) +\sum_{k=b}^{c-1}f(k) =\sum_{k=a}^{c-1}f(k).$

This interpretation has the big advantage that many summation formulas hold for positive and negative upper summation index simultaneously.

For your sum, all common interpretations give $0$ even though the last interpretation gives $\sum_{k=0}^{-2}f(k) = -f(-1)$

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    @9325: For noninteger indices and positive increment, one keeps adding the increment as long as the "running total" is less than the ending index, so yeah... (in the case you gave, keep adding `1` to `1/2` as long as the running total's less than `2`.)2011-04-26
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I think it is a convention that summations start at the lower limit and increment by $1$ until the upper bound is reached. In your example there is no allowable value for $k$ that is no larger than the upper bound. You are adding together an empty collection of elements. Suppose that $A$ is a set of numbers. Consider

$N = \sum A$ and

$ \sum (A \cup \varnothing) = \sum A + \sum \varnothing.$

Since $A = A \cup \varnothing$ the two sums should be equal. But then $\sum \varnothing = 0$.

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    @Arturo Magidin: De-orphaned the period. Thank you for the info.2011-04-25