Does $\sum_{r=0}^n 1=n+1$? Here's my question:
Is the following true? $$\sum_{r=0}^n 1=n+1$$
Here is my area of confusion. Looking at this superficially it would indeed seem to be the case that the result is true, but the result contradicts my understanding of what the sum from $r=a$ to $b$ of a constant means.
In the case of
$$\sum_{r=1}^n 1$$ I would make sense of the summand by thinking of it as
$$\sum_{r=1}^n r^0$$
so the summand would still be a function of $r$. But in the case of
$$\sum_{r=0}^n 1$$
I can't think of it as
$$\sum_{r=0}^n r^0$$
as $0^0$ is undefined. So, is the result correct, or is meaningless?
Note: I know that $\lim_{x\to0} x^x=1$, but it doesn't seem right to apply this to our case of $0^0$; I seem to remember a different value that can found for the 'value' of $0^0$ by taking a different limit.
Thank you for your help.
 A: Hint: You can approach the serie as
$$\sum_{r=0}^{n} 1 = 1 + \sum_{r=1}^{n} 1 = 1 + n$$
A: Your question seems to have been answered in the comments, but here is another way to understand it. The symbol
$$
\sum_{r=0}^{n}f(r)
$$
can be given a precise definition in the following way. First we define
$$
\sum_{r=0}^{0}f(r):=f(0) \, ,
$$
and second for $n\in\mathbb{Z^+}$ we define
$$
\sum_{r=0}^{n}f(r) := f(n) + \sum_{r=0}^{n-1}f(r) \, .
$$
This definition, which references itself in a somewhat strange way, is known as a 'recursive definition'. It might seem kind of circular at first, but if we try unravelling what it means with an example, then we see that there is no issue:
$$
\sum_{r=0}^{2}f(r)=f(2)+\sum_{r=0}^{1}f(r)=f(2)+f(1)+\sum_{r=0}^{0}f(r)=f(2)+f(1)+f(0) \, .
$$
You might want to try proving by induction that
$$
\sum_{r=0}^{n}f(i)=f(0)+\ldots+f(n) \, .
$$
Here, we are dealing with the function $f$ such that for all $x$, $f(x)=1$. Hence,
\begin{align}
\sum_{r=0}^{n}f(i) &= f(0)+\ldots+f(n) \\
&= \underbrace{1+\ldots+1}_{\text{$n+1$ terms}} \\
&= n+1 \, .
\end{align}
One other thing you said in your question is that the summand has to be a 'function of $r$'. This kind of phrasing is very common in mathematics, but is confusing, in my opinion. The reason for this is that the letter used  to represent a function is irrelevant. If I let $f$ be the function defined by $f(x)=x^2$ for all $x$, then this simply means that the output is always the square of the input. The letter $x$ is just being used to tell us this fact in a concise way. Bear in mind that technically $f$ is the function, whereas $f(x)$ is the value of the function $f$ at the point $x$. This subtle point is almost always glossed over in A-level mathematics, but becomes more important later on.
