Generalising the Geometric Series

Level 0: Geometric Series

A series that looks like the following:

$$\frac{1}{c}+\frac{1}{c^2}+\frac{1}{c^3}+\cdots=\sum_{k=1}^{\infty}\frac{1}{c^k}, \quad c \text{ is a constant with } |c|>1$$

is a convergent geometric series, which can be evaluated using the following infinite sum formula:

$$\frac{a}{1-r}$$

where $a$ is the first term of the series and $r$ is the common ratio. In the case of the series above, $a = \frac{1}{c}$ and $r = \frac{1}{c}$. Plugging them in, we obtain the sum, $S_0$, for the above series:

$$S_0=\frac{1}{c-1}.$$

This series is useful in various applications, from distance calculation in physics, compound interests in finance, population growth in biology to even fascinating topics like fractal geometry. Indeed, this series is good and fun, but like most mathematicians may ask, "Can we generalise it even further, and how?"

This article demonstrates one way this series can be generalised, which is by allowing variables in the numerator of the summed over terms.

Level 1: Arithmetico-geometric Series

Now, what if instead of the numerator being a constant number, it is the index of summation $k$?

$$\sum_{k=1}^{\infty}\frac{k}{c^k}, \quad c \text{ is a constant with } |c|>1.$$

It somewhat resembles the previous series but is no longer a geometric one. In fact, it is now an arithmetico-geometric series, with each term in the sum being the product of its respective term of an arithmetic sequence and that of a geometric sequence.

It seems like the infinite sum formula above is not really useful currently. How should we tackle this problem then?

Scaling and Shifting

Since the variable $k$ in the numerator is of degree $1$, let's call the desired sum $S_1$, and list the first few terms of the series.

$$S_1=\frac{1}{c}+\frac{2}{c^2}+\frac{3}{c^3}+\cdots$$

Next, let's multiply all of the terms by $\dfrac{1}{c}$. We then obtain $\dfrac{1}{c} S$, with the terms becoming the following.

$$\frac{1}{c}S_1=\frac{1}{c^2}+\frac{2}{c^3}+\frac{3}{c^4}+\cdots$$

Now, observe that with the exception of the $k=1$ term, there is a difference of $\dfrac{1}{c^k}$ between each term in the former and latter series. With this in mind, let's see what happens if we subtract the former sum by the latter.

$$\frac{c-1}{c}S_1=\frac{1}{c}+\frac{1}{c^2}+\frac{1}{c^3}+\cdots=\sum_{k=1}^{\infty}\frac{1}{c^k}=S_0$$

We recover the geometric series at the beginning! The key here is that we have managed to reduce the problem down to what we've seen before, and we can now apply the geometric infinite sum formula!

How can we express this in terms of $S$ then? Recalling what we have previously done, we can in fact write it as $S-\frac{1}{c} S$. Doing some simplification and applying the infinite sum formula gives us the following equation:

$$\frac{c-1}{c}S_1=\frac{1}{c-1}$$

Solving this equation finally gives us the desired answer, also known as the Gabriel's staircase:

$$S_1=\frac{c}{(c-1)^2}.$$

note

The infinite sum formula has an application in probability theory: it gives us the expected value of a discrete random variable defined by a geometric distribution.

Explanations of some jargons involved.

• A random variable is a representation of a collection of possible outcomes of an experiment, each of which being assigned a probability value.

  There are two types of random variables: discrete and continuous. Roughly speaking, a discrete random variable is "defined over whole (natural) numbers like 1, 2 and 3" whereas a continuous random variable is "defined over decimals (real numbers) like 0.2, 1.67 or even π." Formally, a discrete random variable takes on a countable set of possible outcomes while a continuous random variable takes on an uncountable set of possible values.

  A common example of a discrete random variable, labelled here as $X$, is the roll of a six-sided fair die 🎲. There are six possible results of a die roll: $1,2,3,4,5,6$, each of which occurring with a probability of $\frac{1}{6}$, thus $X=\{1,2,3,4,5,6\}$, and each of the probabilities can be tabulated as follows:

$k$	$P(X=k)$
$1$	$\frac{1}{6}$
$2$	$\frac{1}{6}$
$3$	$\frac{1}{6}$
$4$	$\frac{1}{6}$
$5$	$\frac{1}{6}$
$6$	$\frac{1}{6}$

• The expected value, also called expectation, of a random variable is the average value of all possible outcomes of an experiment, weighted by their respective probabilities. Its formula is

  $$E[X]=\sum_{k \in X}k \cdot P(X=k)$$

  which is the sum of the products of all possible values of a random variable $X$ and their respective probabilities.

  Using the previous die 🎲 example, the expected value obtained from a single die roll is then

  $$1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + 3 \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} + 5 \cdot \frac{1}{6} + 6 \cdot \frac{1}{6} = \frac{7}{2} = 3.5.$$

• A geometric distribution is a probability distribution that gives the number of binomial trials (i.e. you either "succeed" or "fail") needed until achieving the first success.

  If the probability of success on each trial is $p$, then the probability of having the $n$-th trial as the first success is

  $$P(X=n)=(1-p)^{n-1}p.$$

  Using this result, the expected value formula and the fact that the number of trials can theoretically be infinite, the expectation of this distribution is then found to be

  $$E[X]=p \sum_{k=1}^{\infty}{k \cdot (1-p)^{k-1}}$$

  which explains why it is related to the arithmetico-geometric series discussed in this section. In fact, this infinite sum corresponds to the formula with $c$ substituted by $\frac{1}{1-p}$ (so that it lies within the interval of convergence), less the multiplicative constant $p$ and the index shift.

  Upon further simplification while accounting for the multiplicative constant, as well as the index shift by adding a factor of $\frac{1}{1-p}$, we obtain this surprisingly simple result:

  $$p \cdot \cfrac{\frac{1}{1-p}}{\left(\frac{1}{1-p}-1\right)^2} \cdot \frac{1}{1-p}=\frac{1}{p}.$$

This is helpful in calculating the average number of trials needed before attaining the first success. Here are a few examples.

• 🪙 For the $c = 2$ case, the infinite sum, which is also $2$, gives us the expected value of the number of tosses of a fair coin until obtaining the first "tail".
• 🕹️ This formula is also involved in calculating the expected number of encounters required to catch all Pokémons in a simplified, idealisitic Pokémon game, as explained in this Numberphile video featuring Tom Crawford.

note

For the $c=10$ case, this sum also provides a proof that the infinite sum $0.1+0.02+0.003+0.0004+\cdots=\sum_{k=1}^{\infty}\frac{k}{10^k}=\frac{10}{81}$ is in fact rational, which may not be obvious at first glance.

warning

This method only works if the series is indeed convergent. In other words, it won't work for a series like 1 - 1 + 1 - 1 + ... (which has an oscillating sum) or 1 + 1/2 + 1/3 + ... (which, surprisingly, blows up to positive infinity).

As a sidenote, this also means that the use of the same method covered in the infamous Numberphile -1/12 video to "evaluate" the sum of natural numbers is actually wrong (as explained with more detail in this Mathologer video), because the series mentioned in that video are clearly not even convergent to begin with!

A proof to show that $\sum_{k=1}^{\infty}\frac{k}{c^k}$ is indeed convergent.

The ratio test will come in handy here, i.e. we need to evaluate the limit $L = \lim\limits_{k \to \infty}\left\vert{\frac{k+1\text{-th term}}{k\text{-th term}}}\right\vert = \lim\limits_{k \to \infty}\left\vert{\cfrac{\frac{k+1}{c^{k+1}}}{\frac{k}{c^k}}}\right\vert$, which gives us $L=\frac{1}{c} < 1$, so the series is convergent.

In fact, it exhibits a stronger form of convergence: it is absolutely convergent, i.e. even if we replace each term in the series with the absolute value of themselves, the series will still converge!

Level 2: Quadratic-geometric Series

At this point, one may ask, "Why stop here? Why not replace the numerator from $k$ to $k^2$ and find the general formula of the resulting new series? It will still be convergent (proof below) anyways."

$$\sum_{k=1}^{\infty}{\frac{k^2}{c^k}}$$

Let's try to apply the same technique we used to obtain the general formula for the case when $p=1$. Here, we call the desired sum $S_2$.

$$S_2=\frac{1}{c}+\frac{4}{c^2}+\frac{9}{c^3}+\cdots$$

Multiplying throughout by $\dfrac{1}{c}$ yields

$$\frac{1}{c}S_2=\frac{1}{c^2}+\frac{4}{c^3}+\frac{9}{c^4}+\cdots$$

Finally, subtracting the former by the latter, we obtain

$$\frac{c-1}{c}S_2=\frac{1}{c}+\frac{3}{c^2}+\frac{5}{c^3}+\cdots=\sum_{k=1}^{\infty}\frac{2k-1}{c^k}=2S_1-S_0.$$

Once again, we've managed to utilise the fact that the pairwise differences of consecutive perfect squares produce the sequence of odd numbers to reduce this series down to simpler results.

Applying the previous results obtained so far, we finally have

$$\frac{c-1}{c}S_2=\sum_{k=1}^{\infty}\frac{2k-1}{c^k}=2 \cdot \sum_{k=1}^{\infty}\frac{k}{c^k}-\sum_{k=1}^{\infty}\frac{1}{c^k}=2 \cdot \frac{c}{(c-1)^2}-\frac{1}{c-1}$$ $$\implies S_2=\frac{c(c+1)}{(c-1)^3}.$$

Level 3: Cubic-geometric Series

Now, what can we get if the numerator is a cubic term?

$$\sum_{k=1}^{\infty}{\frac{k^3}{c^k}}$$

We will call this sum $S_3$ and let's try the same method again.

$$S_3=\frac{1}{c}+\frac{8}{c^2}+\frac{27}{c^3}+\cdots$$ $$\implies \frac{1}{c}S_3=\frac{1}{c^2}+\frac{8}{c^3}+\frac{27}{c^4}+\cdots$$ $$\implies \frac{c-1}{c}S_3=\frac{1}{c}+\frac{7}{c^2}+\frac{19}{c^3}+\frac{37}{c^4}+\cdots \tag{\#}$$

Notice that in (#), the numerator terms are in the form of $k^3-(k-1)^3$, with $k = 1,2,3,\ldots$ This simplifies to $3k^2-3k+1$, which means that (#) can be rewritten as:

$$\frac{c-1}{c}S_3=3\sum_{k=1}^{\infty}{\frac{k^2}{c^k}}-3\sum_{k=1}^{\infty}{\frac{k}{c^k}}+\sum_{k=1}^{\infty}{\frac{1}{c^k}}=3S_2-3S_1+S_0$$

Finally, we then arrive at:

$$\frac{c-1}{c}S_3=3 \cdot \frac{c(c+1)}{(c-1)^3}-3 \cdot \frac{c}{(c-1)^2}+\frac{1}{c-1}$$ $$\implies S_3=\frac{c^3+4c^2+c}{(c-1)^4}.$$

You may start to notice a pattern here, and this will be discussed in the next section.

Level p: Polynomial-geometric Series

All of these naturally lead to the question: what happens if we generalise the power of $k$ on the numerator to be any positive integer, i.e. what is the infinite sum of the following series?

$$\sum_{k=1}^{\infty}\frac{k^p}{c^k}, \quad c \text{ is a constant with } |c|>1,\; p>1 \text{ is an integer}.$$

We know from previous sections that such series are in fact convergent, so we could try to find the respective general formulae for them.

A proof to show that this series is convergent.

Using the ratio test again like the previous proof, we find that $L=\lim\limits_{k \to \infty}\left\vert{\cfrac{\frac{(k+1)^p}{c^{k+1}}}{\frac{k^p}{c^k}}}\right\vert=\frac{1}{c} < 1$ as well, so it is indeed convergent.

Let's call this sum $S_p$. Inductively, one can see that if we obtain $\dfrac{c-1}{c}S_p$, its numerator terms are then in the form of $k^p-(k-1)^p$, where $k=1,2,3,\ldots$

Not only is this where the binomial theorem comes into play, you can observe that the highest-power term, $k^p$ gets cancelled out while the terms of lower powers remain intact. This means that this method enables us to obtain a recurrence relation of $S_p$ in terms of the lower-power sums, i.e. $S_{p-1}$, $S_{p-2}$ and so on, and this can help us to evaluate its general formula.

Since $k^p-(k-1)^p=k^p-\sum_{i=0}^{p}\binom{p}{i}k^{p-i}(-1)^i=\sum_{i=1}^{p}\binom{p}{i}k^{p-i}(-1)^{i+1}$, we then arrive at this recurrence relation:

$$S_p=\frac{c}{c-1} \cdot \sum_{i=1}^{p}(-1)^{i+1}\binom{p}{i}S_{p-i}.$$

In fact, this derivation is very similar to that in this short paper by Alan Gorfin.

A Segue into Combinatorics

Interestingly, the closed form of this general formula has connections with combinatorics. According to a short paper by Tom Edgar, $S_p$ has the following formula, due to Carlitz:

$$S_p=\frac{c \cdot A_{p}(c)}{(c-1)^{p+1}}$$

where $A_p(c)=\sum_{k=0}^{p-1}A(p,k)c^k$ denotes the $p$-th order Eulerian polynomial (following the Wikipedia notation).

The coefficients of the Eulerian polynomial, denoted $A(p,k)$ as above, are called Eulerian numbers. We can construct a triangle similar to the Pascal's triangle using Eulerian numbers, with $p$ starting from $1$ and incrementing one step at a time when traversing down the column, whereas $k$ ranges from $0$ to $p-1$ as we go from the left to the right of each row. This triangle is then usually called the Euler's triangle.

$$\begin{array}{lc} p=1: & 1 \\ p=2: & 1 \quad 1 \\ p=3: & 1 \quad 4 \quad 1 \\ p=4: & 1 \quad 11 \quad 11 \quad 1 \\ p=5: & 1 \quad 26 \quad 66 \quad 26 \quad 1 \\ p=6: & 1 \quad 57 \quad 302 \quad 302 \quad 57 \quad 1 \\ & \quad \quad \quad \vdots \quad \quad \quad \end{array}$$

Eulerian numbers have an application in combinatorics. To quote Wikipedia,

the Eulerian number is the number of permutations of the numbers $1$ to $n$ in which exactly $k$ elements are greater than the previous element (permutations with $k$ "ascents").

For example, $A(3,1) = 4$, gives us the number of permutations of $1,2,3$ with exactly one element being greater than the previous element, i.e. $132, 213, 231, 312$.

This means that this extension of the geometric series provides us a connection to a combinatorial pattern! Isn't that cool?

Level s: Complex-geometric Series

As a sidenote, what if we generalise even further, so that the numerator variable could take on powers of any complex number?

This brings us beyond the realm (heh) of real numbers and opens ourselves up to the world of Dirichlet series and polylogarithm function, which are common research topics in analytic number theory.

A Dirichlet series is defined as a series of the form

$$\sum_{k=1}^{\infty}{\frac{a_k}{k^s}}$$

where $s$ is a complex number and $a_k$ is a complex sequence.

The polylogarithm function is then a particular collection of Dirichlet series for which $a_k$ is a power series (the free variable in the terms of the series has successively increasing integer powers) in $z$, where $|z|<1$ and $z$ is allowed to be complex (hence the absolute value here is in fact the modulus). Here, $s$ is called the order or weight of the function.

$$\operatorname {Li} _{s}(z)=\frac{z}{1^s}+{z^{2} \over 2^{s}}+{z^{3} \over 3^{s}}+\cdots=\sum _{k=1}^{\infty }{z^{k} \over k^{s}}$$

One can then notice that $S_p$ discussed previously is in fact $\operatorname{Li}_{-p}\left(\frac{1}{c}\right)$, with $|c| > 1$.

Unfortunately, even allowing analytic continuation, obtaining a value or an explicit expression for $\operatorname{Li}_{-s}\left(\frac{1}{c}\right)$ for even just positive rational values of $s$, let alone a non-real number, is difficult. For example, even a closed-form expression for $\operatorname{Li}_{-\frac{3}{2}}\left(\frac{1}{c}\right)=\sum_{k=1}^{\infty}\frac{k\sqrt{k}}{c^k}$ is currently unknown.

That said, polylogarithm function is still a topic with various directions of research, including its integral representations, dilogarithms (i.e. when $s = 2$) and polylogarithm ladders.

To read more about the subject, as well as other topics covered here, feel free to search online using the keywords or navigate to the references/further reading section below.

References, further reading

• Arithmetic-Geometric Progression - Brilliant Wiki
• Arithmetico-geometric series - Art of Problem Solving Online
• Edgar, T. (2018). Staircase Series. Mathematics Magazine, 91(2), 92–95.
• Gorfin, A. (1989). Evaluating the Sum of the Series $\sum_{k=1}^{\infty}{\frac{k^j}{M^k}}$. The College Mathematics Journal, 20(4), 329–331.
• Polylogarithms - Digital Library of Mathematical Functions
• Lewin, L. (1981). Polylogarithms and Associated Functions.
• Questions on Math Stack Exchange with Polylogarithm Tag