Validity of geometric series formula for $r=0$ I can convince myself of the geometric series formula
$$\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$$
for $0<|r|<1$, but not for $|r|<1$ because I don't believe the formula for $r=0$.
If $r=0$, we have
$$\sum_{n=0}^{\infty} r^n = 0^0 + 0^1 + 0^2 + \ldots$$
It is not clear to me what this sum equals, much less that it equals $\frac{1}{1-0}=1$. However, every source that I've consulted says that the result holds for $-1<r<1$.
Can anyone justify the $r=0$ case? Must we simiply accept $0^0=1$ in this context?
 A: In this context, $0^0=1$. Therefore, the sum is $1$.
A: Power series come up everywhere in mathematics, necessitating a convenient form to represent them. The easiest form is
$$
\sum_{k=0}^\infty a_k (x-x_0)^k
$$
In order for this to represent a proper function, we should be able to substitute any value of $x$ into it. If you do not accept the convention $0^0=1$, you then run into problems when $x=x_0$; the value of the power series at that point is supposed to be $a_0$, but you instead get it is $a_0\cdot 0^0$. To avoid this, you would have to instead write
$$
a_0+\sum_{k=1}^\infty a_k (x-x_0)^k
$$ 
which is inconvenient. Therefore, for ease of notation, we stipulate that $0^0=1$ in the context of power series. This is the context of $\sum_{n\ge 0}r^n$. See 
https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero#Polynomials_and_power_series 
for a confirmation of this.

As a side note, there are an overwhelming number of situations where it is convenient to define $0^0=1$, and there are no situations where it is convenient to assume otherwise. 
A: Note a geometric sequence is defined in general as being $\{a, ar, ar^2, ar^3, \ldots \}$, i.e., where each term is $t_i = ar^i$ for $i \ge 0$.
Your statement of $\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$ is actually a specific case of the more general one, such as provided at Geometric series: Formula of
$$\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} \tag{1}\label{eq1}$$
where $|r| \lt 1$, and in your case $a = 1$. As such, if $r = 0$, then the geometric sequence would be $\{1, 0, 0, 0, \ldots \}$ and, thus, it's clear that the sum is $1$. Plugging $a = 1$ and $r = 0$ into \eqref{eq1} gives this same result. Also, by the definition of the sequence, it needs to use "$0^0 = 1$" in the LHS of \eqref{eq1} to get that the first term is $a$. This is due to, for $r \neq 0$, that $r^0 = 1$, so $\lim_{\, r \to 0}r^0 = 1$.
Note that some definitions of geometric sequences requires that $r \neq 0$. However, as you can see, the general equation can still work even if you use $r = 0$.
