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For expressions in the form:

$\sum_{i=1}^{k}f(i),$

does this preclude the possibility that $k$ can be non positive as well as non-integer?

Or more explicitly, can I make an induction on $k$? and take $1$ as base case? And then show that it is true for $k$ if it is true for $k-1$ ?

Thank You.

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    Not necessarily. Some authors define such sums so they obey laws [analogous to (in)definite integrals,](http://en.wikipedia.org/wiki/Integral#Conventions) so one gets additivity on intervals, with an analogous discrete law for reversing the limits of summation.2012-11-23

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This format does preclude non-integer numbers. It specifies that $i$ takes the integers between $0$ and $k$ inclusive, so yes, you could induct on $k$.

If we want to take non integer values, we can do that by specifying an "indexing set" $I$ and saying that $I$ contains all the values we want. We then write $\displaystyle \sum_{i \in I}f(i)$ to say that $i$ takes all values in the set $I$. This a commonly used in, for example, probability, where we take $I$ to be the set of all possible events and $f(i)$ the probability that $i$ occurs.

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For expressions of the form

$\sum_{i=1}^{k}f(i) = \text{____}$

This does this preclude the possibility that $k$ can be non-positive and it precludes non-integer values of $i$. (That is, $i\in \mathbb{Z},\;0\leq i \leq k$ in your expression).

Yes, you can make an induction on $k$, taking $1$ as your base case and showing that if we take $k-1$ to be true, then $k$ is also true.

Note: I used the $\text{____}\;$ to denote some expression because to prove anything about your sum, you need to prove some claim about the sum: e.g. $\sum_{i=1}^{k}f(i) = \text{____}\;$ or that $\;\sum_{i=1}^{k}f(i) \leq \text{____}\;$.

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    @amWhy: Even some nice dialogue +12013-05-16