Is there any good reason that in educational materials, I consistently see the formula for calculating geometric series in canonical form as: $$\sum_{k=0}^{n-1} ar^k = a \frac{1-r^n}{1-r}$$

While an equally correct version is given by [edit: I have now found a book that uses this form] $$\sum_{k=0}^{n-1} ar^k =a \frac{r^n-1}{r-1}$$

The canonical form seems to be backwards for the understanding of most students, as initial examples usually have $r > 1$.

A case of special interest is the infinite series with $r<1$, with the limit of $$\sum_{k=0}^{\infty} ar^k =a\frac{1}{1-r} \text{ (when r<1)}$$

A geometric sequence is simply the discrete version of the exponential function, so a geometric series is simply the discrete version of the integral of the exponential function.

But I can't imagine that anyone would write [edit: an answer below shows that accountants do use it in this form - though I would think it is motivated by usefulness rather than consistency with the infinite sum]: $$\int_{0}^{n} ar^x dx = a\frac{1-r^n}{-\ln r} + C$$ Just because a particular definite integral is of the form: $$\int_{0}^{\infty} ar^x dx = a\frac{1}{-\ln r} \text{(when r<1)}$$

As a side note, I've never seen integration of exponentials taught with regard to geometric series, even though they are normally taught in close proximity, and the formulae are nearly identical (x-1 is a good approximation of ln x for x close to 1). Has anyone had experience with this?

  • $\begingroup$ You don't see the formulas written in either way, since the sum starting at $k=1$ is not given by such formulas, even in the case of an infinite series. $\endgroup$
    – KCd
    Commented Mar 11, 2015 at 14:06
  • 2
    $\begingroup$ Maybe the question should be why "initial examples usually have $r>1$." I seem to encounter the $r<1$ case more often, but I realize that might anomalous. $\endgroup$ Commented Mar 11, 2015 at 14:06
  • $\begingroup$ It would help if you provided some evidence for your claim to "consistently" see the formula one way rather than another. Can you pick 10 random books with the formula in it and say the clear majority have it the first way? $\endgroup$
    – KCd
    Commented Mar 11, 2015 at 14:08
  • $\begingroup$ @KCd Thanks for catching my typo. As for evidence, I assume your "10 random books" was intended as humorous? With the books that I have available and looking at a fair representation of educational websites I did not find a single example of the alternative formula. I consider this to be significant enough to back up my observation that the former formula is the dominant form. $\endgroup$
    – Richard
    Commented Mar 11, 2015 at 14:24
  • $\begingroup$ @AndreasBlass I rarely encounter either in real life, but the bank interest example is a common introduction to geometric series. In my experience students are generally more comfortable with multiplication by factors greater than 1. It could be that most examples are in fact r<1. Are most educational applications also r<1? $\endgroup$
    – Richard
    Commented Mar 11, 2015 at 14:31

5 Answers 5


I agree that the form $\frac{r^n-1}{r-1}$ is in many respects more natural: certainly it is in cases where $r>1$, but even when $r$ is left unspecified, I think this form arises most naturally from many "customary" ways of deriving the formula.

So why is it usually taught the other way around? I suspect the answer is precisely that, in most secondary curricula that I am familiar with, as soon as we have learned the formula for summing a (finite) geometric series we immediately proceed to infinite geometric series: we restrict attention to the case where $\vert r \vert <1$ and say that as $n \to \infty$ the $r^n$ term can be neglected. If we wrote the formula out the way you want to, we would then end up with $\frac{-1}{r-1}$ as the sum, which is weird and confusing because it has an explicitly negative numerator but only an implicitly negative denominator (that is, $r-1$ doesn't look obviously negative, and only is seen to be so if you think about the size of $r$), and so it seems to the novice as if we are saying the infinite series has a negative sum.

Much better and less confusing to write the sum of the infinite series in the form $\frac{1}{1-r}$, and that is the reason, I think, for what you refer to as the "canonical" form: it provides cohesion between the finite and infinite cases.

  • $\begingroup$ I agree that the link to geometric series is the only compelling reason to consider writing the finite geometric series formula in the "weird" way. For a finite geometric series I'd write $1 + r + \dots + r^n$ as $(r^{n+1}-1)/(r-1)$. $\endgroup$
    – KCd
    Commented Mar 11, 2015 at 20:55

As a financial practitioner where such calculations are commonplace, I would encourage you to teach both and motivate them.

In finance an annuity is a stream of regularly spaced payments. An "ordinary" annuity, e.g. a mortgage, has the first payment incurred at the beginning of the first period but paid at the end of the first period, so it includes interest.

The present value of an ordinary annuity, with interest rate $i$, using the handy actuarial notation, is written

$$\require{enclose}a_{\enclose{actuarial}{n}i} = (1+i)^{-1}+(1+i)^{-2}+...+(1+i)^{-n} = \frac{1-(1+i)^{-n}}{i}$$

But it is common for actuaries to use the convention $v = (1+i)^{-1}$ and write $\require{enclose}a_{\enclose{actuarial}{n}i} = \frac{1-v^n}{i}$ instead.

The future value is generally written

$$\require{enclose}s_{\enclose{actuarial}{n}i} = (1+i)^{n-1}+(1+i)^{n-2}+...+(1+i)^{0} = \frac{(1+i)^n-1}{i}$$

Or, alternatively $\require{enclose}s_{\enclose{actuarial}{n}i} = \frac{v^{-n}-1}{i}$

These formulas have a natural symmetry and the relation $\require{enclose}a_{\enclose{actuarial}{n}i} (1+i)^n = s_{\enclose{actuarial}{n}i}$ links these two together.

The present value calculation is common for amortizing, e.g. determining level payments for a loan, while the future value calculation is common for determining level premiums or savings for a future payout, e.g. retirement goal.

Note that present value for continuous case is analogously written

$$\require{enclose}\bar{a}_{\enclose{actuarial}{n}} = \int_0^n v^t dt = \frac{1-v^n}{\delta}$$

where $\delta = \ln(1+i)$ is called the force of interest, i.e. the rate of accumulation is $a(t) = e^{\delta t}$.

So you do generally find the formula expressed in a similar fashion to the "I can't imagine that" form you provided, albeit with a different convention for variables.

As far as derivation is concerned, the most natural seeming to me is to teach the perpetuity

$$\require{enclose}a_{\enclose{actuarial}{\infty}} = v + v^2 + ...$$

Where the trick is to see that

$$\require{enclose}a_{\enclose{actuarial}{\infty}} = v + v a_{\enclose{actuarial}{\infty}}$$

may be used to derive

$$\require{enclose}a_{\enclose{actuarial}{\infty}} = \frac{v}{1-v} = \frac{1}{i}$$

and the relationship

$$\require{enclose}a_{\enclose{actuarial}{\infty}} = a_{\enclose{actuarial}{n}} + v^n a_{\enclose{actuarial}{\infty}}$$

may be used to derive

$$\require{enclose}a_{\enclose{actuarial}{n}} = (1-v^n) a_{\enclose{actuarial}{\infty}} = \frac{1-v^n}{i}$$


The first example of this I saw was the sum of the divisors of a prime power $p^n$: $$ 1+p+p^2+\dots+p^n = \frac{p^{n+1}-1}{p-1} $$ and in that case $p>1$ is certainly intended. Then it seemed pretty strange to me when my teacher wrote it as $$ \frac{1-p^{n+1}}{1-p} $$ as part of his explanation of the formula. Only later in calculus did I see that this was the natural way to write it when working on infinite series.

  • $\begingroup$ Out of interest, how did you come to investigate the sum of divisors of prime powers? I haven't seen that example before. Could not find much online, just this oeis.org/A023194 Numbers n such that sigma(n) (sum of divisors of n) is prime - all of which happen to be prime powers. $\endgroup$
    – Richard
    Commented Mar 11, 2015 at 22:03
  • $\begingroup$ It is even in Euclid, when he talks about perfect numbers. $\endgroup$ Commented Mar 12, 2015 at 14:24
  • $\begingroup$ @Richard en.wikipedia.org/wiki/Divisor_function $\endgroup$ Commented Jun 6, 2022 at 16:10

It may have something to do with the classical proof of the formula: $$a(1+r+\dots+r^{n-1})(1-r)=a(1-r+r-r^2+\dots+r^{n-1}-r^n)=a(1-r^{n+1})$$ which makes slightly more sense this way given that we usually write the sum $\sum ar^k$ in increasing term order.

  • $\begingroup$ It is also related to a proof using a special case of a generalised binomial theorem en.wikipedia.org/wiki/… - $\frac{1}{(1-x)^s}$ with $s=1$. Though I don't think this is a common derivation, I think the influence of binomial notation may reinforce the order. $\endgroup$
    – Richard
    Commented Mar 11, 2015 at 22:16

Practical problems for this formula that involve 0 < r < 1 are more common than similar problems that involve r > 1.

Mortgage interest calculations (such as present value interest factor of an annuity) involve 0 < r < 1. These problems involve a known (or infinite) n = number of terms.

Geometric expansion problems involving r > 1 do exist, but the problem formulator must arbitrarily decide on n = number of terms. In order to have a finite answer, there must be a limit to the number of terms -- but it is often not obvious how such a limit should be chosen.


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