# How to Recognize a Geometric Series

What is the "definitive" definition of a geometric series?

I phrased the question this way, because I've checked multiple Calculus textbooks, as well as Paul's Online Math Notes, and they seem to give conflicting definitions.

Thomas Calculus (12th ed) says, "Geometric Series are series of the form:" $$\sum_{n=1}^{\infty}ar^{n-1}$$

Calculus for Scientists and Engineers says, "Geometric Series have the form:" $$\sum_{k}^{\infty} r^k \space \text{or} \space \sum_{k}^{\infty}ar^k$$ and that $a$ is simply the first term.

My textbook, Stewart Calculus Early Transcendentals (7th ed) says, "An important example of an infinite series is the geometric series" $$a+ar+ar^2+...+ar^{n-1}+... = \sum_{n=1}^{\infty} ar^{n-1}$$

But then Stewart goes on to provide examples and exercises which do not fit that form, such as: $$\sum_{n=1}^{\infty}ar^n$$

Paul's Online Math Notes say that a geometric series must have one of the forms: $$\sum_{n=0}^{\infty} ar^n \space \text{or} \space \sum_{n=1}^{\infty} ar^{n-1}$$

This was all especially confusing for me while trying to do homework, and in class, when we were shown the following problem:

$$\text{Determine whether the following series converges or diverges, and if it converges, find the sum:} \space \sum_{n=1}^{\infty} \frac{1}{2^n}$$

I recognized the common ratio $r=\frac{1}{2}$, but I thought that $a=1$. It turned out, however, that $a=\frac{1}{2}$, also.

This would have made sense to me if the series had had the form $\sum_{n=1}^{\infty}ar^{n-1}$, but it didn't.

In other words, if the index begins with $n=1$, how can you end up with a series of terms $a+ar+ar^2+ar^3+...$? The first term would be $ar^1=ar$; there's no way you could have $ar^0=a(1)$, when the index begins with 1.

And if that's the case, how can $\frac{1}{2}$ be simultaneously equal to $a$ and to $r$, when there is only a single term in the series definition itself?

Is there a definitive, universal definition for a Geometric Series, or is it somewhat subjective and/or open to interpretation? Or am I simply missing something here?

• $\sum_k^{\infty}ar^k$ is a shorthand notation so that isn't a different definition. The example with $\sum_{n=1}^{\infty}ar^{n}$ is basically same as $\sum_{n=1}^{\infty}ar^{n-1}$, so it makes sense that it was used in an example. $\sum_{n=0}^{\infty}ar^{n}$ , $\sum_{n=1}^{\infty}ar^{n-1}$ are the exact same sum except it is written with different indices. The definition is the same in all of the examples, but the authors just wrote them in different ways. – user222031 Apr 1 '16 at 2:01
• I hate to say this but indexing conventions are considered far too trivial to be worth standardizing. Also it's assumed one will shift indexing at will to solve. – fleablood Apr 1 '16 at 2:06
• Yes, $\sum_{n=0}^{\infty} ar^n$ and $\sum_{n=1}^{\infty} ar^{n-1}$ are the same, with a simple shift in the index. But how can $\sum_{n=1}^{\infty} ar^n$ be the same? In the first two, the first term is raised to the 0th power. In the latter, the first term is raised to the 1st power. Not the same, afaik. – tommytwoeyes Apr 1 '16 at 3:44
• In other words, it wasn't the indexing conventions that tripped me up. It was the conflicting definitions different textbooks, online resources, and my instructor gave which lead to my confusion. – tommytwoeyes Apr 1 '16 at 3:59
• This isn't the question you asked, but to solve the problem you were given, I assume you were using the infinite geometric series formula $\frac{a}{1-r}$, when $|r|<1$. Although you can shift indexing and/or change $a$, you absolutely must use the same definition of $a$ used by the formula. So, it's not the definition, but the definition used by that formula. Note that Stewart's $\sum_{n=1}^{\infty} ar^n$ does define geometric series (your question), just with a different definition of $a$ (your problem). You can convert to that form by using $a_{Stewart} = a/r$ to counter his extra $r$. – hyperpallium Feb 8 at 2:28

The definition of a geometric series is a series where the ratio of consecutive terms is constant. It doesn't matter how it's indexed or what the first term is or whether you have a constant. That stuff just has to do with how you write the series. So the '$a$' in a given series need not be unique, as you pointed out. But the common ratio $r$ is uniquely determined by the series. $$\sum\limits_{n=0}^{\infty} a r^n$$ $$\sum\limits_{n=3}^{\infty} r^n$$ $$\frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots$$ $$1+1+1+ \cdots$$ are all examples of geometric series (the last one doesn't converge). For example, the first one is a geometric series because the ratio of consecutive terms is $$\frac{ar^{k+1}}{ar^k} = r$$

• Thank you. Maybe it's perfectly clear to everyone but me, but the various textbooks, as well as Paul's Online Math Notes, and my instructor, included the format of the series (index, terms, etc) in the definition of Geometric Series. That's what confused me and caused the apparent contradictions. I wish they'd explained it as plainly as you did. Cleared it right up. – tommytwoeyes Apr 1 '16 at 3:50

All the definitions here are equivalent : $$\sum_{i=1}^{\infty}{ar^{i-1}}=ar^0+ar^1+ar^2+...=\sum_{j=0}^{\infty}{ar^j} ~~~~\text{where}~(j = i-1)$$

Moreover if $a = r$, then $ar^0 = a = r = r^1$ so $$\sum_{p=0}^{\infty}{ar^p}=\sum_{q=1}^{\infty}{r^q} ~~~~~\text{where}~(p=q-1)$$

But this is indeed confusing when we are used to finite sum, where, for $M\in \mathbb{N^*}$ $$\sum_{k=1}^{M}{ar^{k-1}} = ar^0+ar^1+...+ar^{M-1}=\sum_{m=0}^{M-1}{ar^m} ~~~~~\text{where}~(m = k-1)$$

$$\frac{1}{2^n}=\left(\frac{1}{2}\right)^n$$ So you could choose $r_1$ to be $\frac{1}{2}$ and write your serie as $$(a_1,r_1)\leftarrow(1,\frac12) ~~~~~\sum_{i=1}^\infty{r_1^i}$$ or choose $a_2$ to be $\frac12$ and write $$(a_2,r_2)\leftarrow (\frac12,\frac12) ~~~~~ \sum_{i=0}^{\infty}{a_2r_2^i}$$
• Yes, thanks. That makes more sense. I was caught up in the language that some of the textbooks used, in which they said something similar to, "A geometric series is a series of the form $\sum_{n=1}^{\infty} \mathbf{ar^{n-1}}$, with a common ratio between terms. I thought that the form was required as part of the definition. Now I understand that it's simply a series which has a common ration between consecutive terms. – tommytwoeyes Apr 1 '16 at 3:55
Shift the index and subtract $$\sum_{n=1}^\infty \frac{1}{2^n} = \sum_{n=0}^\infty \frac{1}{2^n} - 1$$ or, you can multiply and multiply $$\sum_{n=1}^\infty \frac{1}{2^n} = \frac{1}{2}\sum_{n=0}^\infty \frac{1}{2^n}$$