# 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.

Can anyone justify the $$r=0$$ case? Must we simiply accept $$0^0=1$$ in this context?

• if $r=0$ it's not geometric series. By definition, ratio of consecutive terms should be the same. – Vasya Mar 7 at 19:16
• 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. – Thomas Andrews Mar 7 at 19:18
• Then why does every textbook (even good ones, like Spivak) give the formula for $-1 < r <1$? – mathclassfromscratch Mar 7 at 19:19
• If $r=0$ is allowed, the first term can be any number and $0^0=1$ does not help – Vasya Mar 7 at 19:29
• 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 $|r|<1$. – Paramanand Singh Mar 8 at 5:51

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

• Why is $0^0=1$ in this context? Is it different in other contexts? – John Douma Mar 7 at 19:21
• The first paragraph here suggests that context matters: en.wikipedia.org/wiki/Zero_to_the_power_of_zero – mathclassfromscratch Mar 7 at 19:29
• @mathclassfromscratch No, it says there is no agreed upon value for $0^0$. – John Douma Mar 7 at 19:31
• @JohnDouma Yes, and then the second sentence says that context matters. – mathclassfromscratch Mar 7 at 19:36
• @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. – 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.

• What number can be considered overwhelming comparing with zero? – 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 $$r = 0$$.

• 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. – user3329719 Mar 11 at 6:39
• @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. – John Omielan Mar 11 at 7:07