Sum including $0^0$ So if I have the sum
$$\sum\limits_{r=0}^n r^k$$
It makes intuitive sense that for $k=0$ the sum equals $n+1$ but of course in this particular situation the sum includes the term $0^0$ which of course is undefined and here we can't take any limits to find what term we should use as this is a discrete sum.
So the question is, is this sum equal to $n+1$ and if so why?
 A: You would have to consider the function $f(k) = \sum_{r=0}^{n} r^k$ and the domain of $k$. Assuming $k \in \mathbb{R}$, then we take the right limit $k \to 0^+$ and the left limit $k \to 0^-$.
The left limit does not exist, because we cannot define what it means to have (e.g.) $0^{-0.1}$.
The right limit converges to $n$, because we have (e.g.) $0^{0.1} = 0$.
So you may use the right limit to define $\sum_{r=0}^{n} r^0 = n$. Though you have to mention that you got it through the limit.

P.S: Considering that the term $r=0$ does not contribute to the sum in anyway for positive k, and it is not defined for negative k,why put it in the sum?
A: Usually, the exponentiation operation in situations like this should not be interpreted as the usual operation on real numbers.
There are a couple of interpretations of what is really meant here by the notation $r^k$, but they generally have the property that substituting $k \mapsto 0$ results in the constant $1$. So, further substituting $r \mapsto 0$ still produces the value $1$.
A: Contrary to your saying, one can consider
$$\lim_{k\to0}\sum_{r=0}^n r^k=\sum_{r=0}^n\lim_{k\to0} r^k=n$$
because $\forall k>0:0^k=0$.
