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.
On
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 $\;\sum_{i=1}^{k}f(i) \leq \text{____}\;$.
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.