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?

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

In this context, $0^0=1$. Therefore, the sum is $1$.

  • $\begingroup$ 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

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


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.

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

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 $r = 0$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.