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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?

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    $\begingroup$ if $r=0$ it's not geometric series. By definition, ratio of consecutive terms should be the same. $\endgroup$ – Vasya Mar 7 at 19:16
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    $\begingroup$ There are lots of ways to define geometric series, @Vasya. One is that $a_{n+1}a_{n-1}=a_n^2.$ In any event, this nit-pick doesn't resolve the question. $\endgroup$ – Thomas Andrews Mar 7 at 19:18
  • $\begingroup$ Then why does every textbook (even good ones, like Spivak) give the formula for $-1 < r <1$? $\endgroup$ – mathclassfromscratch Mar 7 at 19:19
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    $\begingroup$ If $r=0$ is allowed, the first term can be any number and $0^0=1$ does not help $\endgroup$ – Vasya Mar 7 at 19:29
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    $\begingroup$ Let's say that a correct/umabiguous version of the formula in question is $1+\sum_{n=1}^{\infty}r^n=\dfrac{1}{1-r}$ for $|r|<1$. $\endgroup$ – Paramanand Singh Mar 8 at 5:51
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In this context, $0^0=1$. Therefore, the sum is $1$.

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  • $\begingroup$ Why is $0^0=1$ in this context? Is it different in other contexts? $\endgroup$ – John Douma Mar 7 at 19:21
  • $\begingroup$ The first paragraph here suggests that context matters: en.wikipedia.org/wiki/Zero_to_the_power_of_zero $\endgroup$ – mathclassfromscratch Mar 7 at 19:29
  • $\begingroup$ @mathclassfromscratch No, it says there is no agreed upon value for $0^0$. $\endgroup$ – John Douma Mar 7 at 19:31
  • $\begingroup$ @JohnDouma Yes, and then the second sentence says that context matters. $\endgroup$ – mathclassfromscratch Mar 7 at 19:36
  • $\begingroup$ @mathclassfromscratch The justifications come from different contexts. That doesn't mean that there are provable values for $0^0$ based on different contexts. Either way, I can say this sum equals $\frac{1}{\sqrt{\pi}}$ and there is no context in which you can prove that $1$ is a better answer. $\endgroup$ – John Douma Mar 7 at 19:42
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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.

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  • $\begingroup$ What number can be considered overwhelming comparing with zero? $\endgroup$ – user Mar 7 at 20:20
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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 $r = 0$.

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  • $\begingroup$ This answer basically just sidesteps the problem by defining a geometric sequence informally. You could have also defined it as 1,r,r^2,... and be done with it. The original question basically asks why when we define a geometric series as a_n=ar^n we should have a_0=a0^0 be a rather than anything else. You're just assuming that this is indeed the correct definition from the start. $\endgroup$ – user3329719 Mar 11 at 6:39
  • $\begingroup$ @user3329719 The original definition of a geometric series that I learned, and as also defined by the referenced article, is as I state at the top. Although I define it informally initially, I also define the terms formally as $t_i$. In addition, as for why, in $a_0 = a 0^0$, it makes sense for $0^0 = 1$, I also explain that with my statement about the limit as $r \to 0$, so I'm showing why it makes sense for $0^0 = 1$ instead of assuming this at that time. As such, I'm not clear exactly what issues you have with my answer. $\endgroup$ – John Omielan Mar 11 at 7:07

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