Prerequisites:
In number theory, to understand some proofs, a certain familiarity with the concept of “numerical series” is necessary. In this article, after briefly introducing this concept, we’ll focus on some series used in number theory, which are not always studied at school or university; we’ll see in particular some techniques which are useful in practice to operate with such series.
This article does not claim to treat numerical series in a rigorous nor exhaustive way, but it can be seen as a complementary study for those who have already studied series, and as a starting point for further study for those who have never studied them.
A brief introduction to numerical series
We’ll begin our study of series with a concrete case, which often occurs in number theory. Indicating with p a prime number, we can consider the following expression, which is an example of a numerical series:
Equivalently, one can also use the following notation, which describes the series in question as a summation (it is no coincidence that the Greek letter \Sigma is the equivalent of the letter S, the initial of “sum”), for each i ranging from 1 to infinity, of the inverses of all the prime numbers raised to the power of i:
From now on, however, we’ll prefer the dotted notation, because it is often easier to read. In this notation it is not important how many terms there are before the dots but, the more there are, the clearer it is how the series is composed. In any case, no matter how many terms there are before the dots, the mathematical entity represented, that is the series itself, does not change.
From an intuitive point of view, a series can be considered an “infinite sum” of real numbers; in (1) for example we have the “sum” of the inverses of all powers of p from 1 onwards; the word “sum” is used here in an intuitive and formally improper sense because – it’s good to have this clearly in mind – in mathematics there are no “infinite sums”, but a “sum” of infinite terms is actually defined as the limit of the sequence of its partial sums, where a partial sum means the sum of its first n terms. This limit, calculated as n tends to infinity, is called the sum of the series, even though it is not actually a sum, but a limit of sums. The series we started with, for example, is the limit of the following sequence:
It’s therefore the following limit:
That said, it may be useful to see a series as an infinite sum; the important thing is to keep in mind its true meaning, that is, to remember that it’s the limit of the sequence of its partial sums.
From a graphical point of view, each term of the series can be represented by a rectangle with a base of 1 and a height of the numerical value of the term itself which, since the base is 1, will also coincide with the area. The sum of the series is therefore the limit of the incremental calculation process of the area of the first n rectangles, as n tends to infinity, as shown in the following image which refers to the series (1) for p = 2:
Intuitively, one can think of this limit as the area of the imaginary complete histogram containing infinite rectangles, one for each term of the series; however, it’s good to remember that this “infinite histogram”, like the “infinite sum”, does not exist as a mathematical object, but it’s only a way to see the limit of the graphical process illustrated above.
If the limit of the partial sums of a series exists and is a real number (in particular it’s not \pm \infty), the series is said to be convergent (other types of series, where the limit does not exist or is infinite, are less important for our articles).
The series (1) is convergent, i.e. the limit exists. What guarantees the existence of the limit, based on known results, is that the series is of the kind:
where x is a real number such that |x| \lt 1.
A series like (2) is called geometric series of ratio x, and it’s known that its sum (i.e. \lim_{n \to \infty} \sum_{i = 0}^{n} x^n) is equal to:
In the case of the series (1), being \left| \frac{1}{p} \right| \lt 1, we can set x := \frac{1}{p}, concluding that the sum of the series is:
In particular, the fact that p is prime in this case has no role in making the series convergent; however, the series (1) occurs often in number theory, so much so as to justify the introduction of the following Property:
Series of the inverses of the powers of a prime number
Let p be a prime number. Then:
A technique for calculating the sum of a convergent series
To understand if a series is convergent there are several tests that for brevity we’ll not discuss in this article. Instead, we’ll focus on the numerical aspect, that is, we’ll see how to calculate the sum of a series when it’s already known, by virtue of a certain test or based on some theorem, that it converges. This aspect is useful when studying number theory, in particular with regard to some proofs where there are passages which replace a series with its sum.
In the case of the previous series, we can reason like this:
- Suppose that the series is convergent, that is, suppose that there exists a real number S such that
1 + \frac{1}{p} + \frac{1}{p^2} + \frac{1}{p^3} + \ldots = S
- Let’s multiply everything by p:
p + \frac{p}{p} + \frac{p}{p^2} + \frac{p}{p^3} + \ldots = pS
then let’s simplify:
p + 1 + \frac{1}{p} + \frac{1}{p^2} + \ldots = pS - We note that 1 + \frac{1}{p} + \frac{1}{p^2} + \ldots is the starting series (with one less term before the dots, but, as we said at the beginning, it’s just an alternative representation of the same series); so we can substitute S in place of 1 + \frac{1}{p} + \frac{1}{p^2} + \ldots, obtaining
p + S = pS
- At this point all that remains is to obtain S:
\begin{aligned}p + S = pS & \Rightarrow \\ p = pS - S & \Rightarrow \\ p = (p - 1)S & \Rightarrow \\ S = \frac{p}{p - 1} & \end{aligned}
Formula (3), valid for the series 1 + x + x^2 + x^3 + \ldots, can be obtained with similar steps. The only additional precaution, compared to the previous case, is that we must assume x \neq 1; this hypothesis in fact ensure the validity of the last step of the proof, where we divide by 1 - x.
Suppose that the series converges and has sum S:
Let’s multiply by x:
We can note that the series obtained is the same as the starting series minus the initial term; to find the starting series, we can therefore add that term to both members:
Now we have the starting series on the left, which we can replace with S:
Let’s solve the equation:
Note that in the last step we divided by 1 - x, so we need to add the hypothesis that x \neq 1.
In general, the technique we propose consists of the following steps:
- Assume that the series is convergent, that is, assume that there exists a real number S equal to the sum of the series, then write the equality between the series and S;
- Perform one or more algebraic operations both on the left of the equality (i.e. on the series) and on the right (on S) so that on the left, simplifying, you obtain an expression containing the initial series;
- On the left, substitute S in place of the starting series, obtaining an equation in the unknown S;
- Solve the equation to find the value of S.
This technique has the advantage of making it possible to study numerical series without using other sophisticated techniques, but it also has the defect of not helping us to understand whether a series converges or not; convergence is always a hypothesis that must be verified at the outset.
To underline the importance of the convergence hypothesis, we can try to apply formula (3) to a geometric series that does not converge. For example, if in the series (2) we replace x = 2, we obtain the series 1 + 2 + 4 + 8 + \ldots, which, according to formula (3), would have sum \frac{1}{1 - 2} = \frac{1}{-1} = -1, which is absurd: this is one of the traps of numerical series!
In summary, we must therefore remember that our technique provides the correct result only when it is already known that the series converges, but it does not help to establish whether it converges or not. We will therefore state properties such as the following, in which the hypotheses include both the convergence of the series and any values to be excluded to make the steps of the proof valid:
Power series of a real number
Let x be a real number other than 1. If the series 1 + x + x^2 + x^3 + \ldots converges, then:
In the case of the geometric series, we know that it converges when |x| \lt 1, so in the previous statement we could replace “If the series 1 + x + x^2 + x^3 + \ldots converges” with “If |x| \lt 1“. However, the study of the convergence of a series is beyond the scope of this article; therefore, in the following statements, we’ll be satisfied with a generic hypothesis of convergence.
We can also observe that the hypothesis that x is different from 1 is redundant, because this hypothesis is included in the hypothesis of convergence of the series (if |x| \lt 1, surely in particular x \neq 1). However, the proof does not go into the details of when the series is convergent and when it is not, so within the proof we have no elements to notice the redundancy and, to justify the division step by x - 1, we are forced to add a separate hypothesis.
As regards the verification of the hypothesis of convergence, as we mentioned, there are various tests, which should be dedicated to a separate article. For the moment we limit ourselves to observing that in many cases the non-convergence can be proved simply by calculating the limit of the partial sums: if the limit is infinite, the series does not converge, so we know that we cannot apply the formula. For example, the series 1 + x + x^2 + x^3 + \ldots for x = 2 becomes 1 + 2 + 4 + 8 + \ldots, with partial sums 1, 1 + 2, 1 + 2 + 4, 1 + 2 + 4 + 8, \ldots = 1, 3, 7, 15, \ldots, a sequence that clearly has an infinite limit (it is sufficient to apply the definition of limit, noting that the sequence consists of powers of 2 decreased by one unit).
One thing to keep in mind when applying this technique is that, when performing algebraic operations on series, you should always ask yourself whether these operations are legal. For example, one of the steps we performed to find the sum of the series 1 + \frac{1}{p} + \frac{1}{p^2} + \frac{1}{p^3} + \ldots is the following:
This means that, after multiplying by p each element of the series 1 + \frac{1}{p} + \frac{1}{p^2} + \frac{1}{p^3} + \ldots which by hypothesis has sum S, we’ll obtain a series which has sum pS. This property, which may seem obvious, must actually be proved. The proof should start from the definition of the sum of a series as the limit of its partial sums, then applying the properties of limits. In this case, for example, you should use the property that, if a general term sequence a_n has limit l \in \mathbb{R}, then the sequence with general term c a_n, obtained by multiplying each term of the previous one by a constant c, has limit c \cdot l. For brevity, we’ll omit such kind of details in the proofs of this article; however, we invite our readers to do the exercise of identifying these critical passages and think about which properties of limits can be used to justify them.
If in Property A.20 we substitute \frac{1}{x} for x (with the additional assumption that x \neq 0, since it appears at the denominator), we’ll obtain the more general version of Property A.19:
Series of inverse powers of a real number
Let x be a real number that is equal to 0 or 1. If the series 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \ldots converges, then:
The proof is immediate; in fact, by substituting \frac{1}{x} in place of x in the formula \frac{1}{1-x}, we’ll obtain \frac{1}{1 - \frac{1}{x}} = \frac{1}{\frac{x - 1}{x}} = \frac{x}{x - 1}.
Application examples
The second point of the technique – i.e. the operations to be performed on both members of the equality, in order to obtain on the left member an expression containing the series itself – is the most complicated. In fact, the operations to be performed vary from one series to another and, to find the most convenient operations, you need to have a bit of inventiveness or a bit of experience. In this regard, we propose the following series as further examples that can facilitate the understanding of the mechanism.
Series of the inverse powers of a real number with alternating signs
Let x be a real number that is different from -1 and 0. If the series 1 - \frac{1}{x} + \frac{1}{x^2} - \frac{1}{x^3} + \frac{1}{x^4} - \frac{1}{x^5} + \ldots converges, then:
Since by hypothesis the series converges, its sum is a real number, which we can call S:
Clearly the hypothesis that x is non-zero is necessary, otherwise the terms of the sequence would not be defined.
Let’s multiply by -x:
Let’s simplify:
The series in brackets is the initial one, and by the initial hypothesis we can replace it with S:
Let’s solve the equation:
In the last step we divided by 1 + x, so we used the hypothesis that x \neq -1.
Series of the inverses powers of a real number from the square onwards
Let x be a real number that is not equal to 1 or 0. If the series \frac{1}{x^2} + \frac{1}{x^3} + \frac{1}{x^4} + \frac{1}{x^5} + \ldots converges, then:
Since by hypothesis the series converges, its sum is a real number, which we can call S:
Again, x must be non-zero, so that the terms of the series are defined.
Let’s multiply by x:
Let’s simplify:
The series in brackets is the initial one, and by the initial hypothesis we can replace it with S:
Let’s solve the equation:
In the last step we divided by x - 1, applying the assumption that x \neq 1.
We could have found the sum of this series also using another principle, which consists in referring to a series already known. In fact, the series can be obtained by eliminating the first two terms from the series of inverse powers:
where the sum of the series in parentheses, by Property A.21, is equal to \frac{x}{x - 1}. Substituting, we’ll obtain:
A variant of the technique consists in tracing the series back, through appropriate algebraic steps, to another already known series, different from the initial one. For example, the following series can be traced back to the previous one:
Power series of a real number with increasing coefficients
Let x be a real number different from 1. If the series 1 + 2x + 3x^2 + 4x^3 + 5x^4 + 6x^5 + \ldots converges, then:
Since by hypothesis the series converges, its sum is a real number, which we can call S:
Let’s multiply by (1 - x):
After doing the calculations, we’ll get to:
The series at the first term is that of Property A.20, so we can replace it with its sum, which we already know:
Dividing by 1 - x (which is legal by the hypothesis x \neq 1) we’ll obtain the thesis.
This example could continue indefinitely, obtaining new series that can be traced back to the previous ones already known. For example, the following series can be traced back to the previous one:
Power series of a real number with coefficients equal to the sum of the first integers
Let x be a real number different from 1. If the series 1 + 3x + 6x^2 + 10x^3 + 15x^4 + 21x^5 + \ldots converges, then:
In this series, the n-th term is the sum of the first n integers; for example, 3 = 1 + 2, 6 = 1 + 2 + 3, 10 = 1 + 2 + 3 + 4, etc.
Since by hypothesis the series converges, its sum is a real number, which we can call S:
Let’s multiply by (1 - x):
After doing the calculations, we’ll get to:
But the series at the first term is that of Property A.24, so we can replace it with its sum, which we know:
Dividing by 1 - x, which by hypothesis is different from 0, we’ll obtain the thesis.
From this proof and the previous one, we can grasp what is the characteristic that makes it possible to reduce one series to another: the differences between consecutive coefficients of a series are equal to the coefficients of the previous series. For example, in the last series the differences between the coefficients are 3 – 1 = 2, 6 – 3 = 3, 10 – 6 = 4, 15 – 10 = 5, 21 – 15 = 6, etc.; as you can see, this sequence of differences coincides with the coefficients of the series of Property A.24, from the second term onwards: 2, 3, 4, 5, 6, etc. The same thing is true for the series of Property A.24 with respect to that of Property A.21: the difference between consecutive coefficients of the series of Property A.24 is always 1, just like the coefficients of the series of Property A.21.
This connection between the coefficients of different series is leveraged when a series is multiplied by 1 - x. In fact this operation, as you can see in this proof and in the previous one, has the effect of generating a new series having as coefficients the differences between consecutive coefficients of the initial series; therefore, from the series of Property A.24, we obtained that of Property A.23 and, through the same operation, from the latter we in turn obtained that of Property A.21.
Product of series and order of terms of a series
Sometimes it happens to multiply two convergent series together; for example, assuming that the series 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \ldots and 1 + \frac{1}{y} + \frac{1}{y^2} + \frac{1}{y^3} + \ldots converge, we might want to calculate the product:
To understand what this writing means, we have to remember that, by definition, a series is nothing but the limit of the sequence of its partial sums, so (4) is equivalent to:
By the properties of the limits of sequences, since by hypothesis the two limits in parentheses exist, the following limit also exists and is equal to the product of the two limits:
So, writing the sums in full, the product of the two series is equal to the limit of the following sequence:
Let’s develop each element:
We can note that the k-th element of the sequence is the sum of all terms of the type \frac{1}{x^ny^m}, with 0 \leq n \leq k and 0 \leq m \leq k; consequently, as k increases, sooner or later we’ll obtain all the possible terms given by the product of a term of the first series by a term of the second series, that is, all terms of the type \frac{1}{x^ny^m}, with n \geq 0 and m \geq 0. Often, in cases like this, compact notations such as the following are used:
Notations like this have the merit of being intuitive but the defect of not being rigorous. A summation of this kind, in fact, only says which numbers to add, but not in what order. For example, we could interpret the notation (7) as the sequence of partial sums (6) or as a different sequence, like the following, where the k-th element of the sequence is the sum of all the terms of the type \frac{1}{x^ny^m} with n+m \leq k:
or, alternatively, any other sequence of partial sums that our imagination can suggest, provided that sooner or later all the terms of the series will appear.
For finite sums we know that the order of the addends does not affect the result, but for series this is not a given. How can we be sure, for example, that the series having as partial sums (6) and (8), both of which can be traced back to the form (7), have the same sum?
The question is not at all trivial, but fortunately there is a simple answer: if a series converges and has only positive terms, then indeed the order of the terms does not matter. In more formal terms:
Series convergence and term ordering
Let \sum_{i = 0}^{\infty} x_i be a convergent numerical series with positive terms, having sum S. Then any other series \sum_{i = 0}^{\infty} y_i such that \{x_i \mid i \geq 0\} = \{y_i \mid i \geq 0\} converges and has sum S.
Usually, between one partial sum and the next one, the difference is a single term, while in (6) and (8) we can see that several terms are added simultaneously. This is not a problem, because you can create a new sequence that adds only one term at a time by completing the missing steps; for example, starting from (8) you can create the sequence:
In this case we say that the previous sequence (8) is a subsequence of the new sequence (8′). A theorem on sequences guarantees that, if a sequence has a certain limit, then all its subsequences have the same limit; in our case, if (8′) tends to a limit S, which is the sum of the series (7), the sequence (8) which has some “joined” terms, also tends to S. So it it’not important if in partial sums we add more terms at a time, because this does not change the limit of the sequence.
For a series that includes negative terms, the thesis of the previous Theorem may not hold: it’s possible that two convergent series with the same terms, but taken in a different order, have different sums. For this reason, it’s wrong, for a series of this kind, to conceive it as an “infinite sum” or graphically as an “infinite histogram”: these concepts, being intrinsically infinite, therefore including the effect of all the terms simultaneously, would lead us to think that the sum, or the area of the histogram, are independent of the order of the calculations performed. Actually, the only truly infinite mathematical object is the limit process, which however never considers all the terms simultaneously, but only a finite number of them at any given moment; for this reason in some cases the final result depends on the order in which terms are taken. For further information on this topic, see Theorem 29.7 in the book Foundations of Mathematical Analysis (R. Johnsonbaugh, W. E. Pfaffenberger).
Theorem A.1 justifies the use of notations like (7), where the order of the terms is not specified, when they are all positive, and from the context we know that the series converges.
As an example of application, let’s study the following product of series:
For example, for d = 6 we’ll get:
The numbers we see at the denominator have in common that in their factorization only the prime factors 2 and 3 appear, which are the prime factors of 6, possibly repeated. This is a consequence of how they were obtained, each as the product of an element of the first series by one of the second.
In general, the following Property holds:
Product of the series of inverse powers of the prime factors of a positive integer
Let d be a square-free positive integer. Then:
In this case, each series of the product on the left side converges by Property A.19, and their product converges by the properties of the limit of convergent sequences product, as we saw before in the case of two sequences (if the sequences are more than two, we can just iterate the procedure). On the right, however, we have a series where the order of terms is not specified, but we know that it’s not important because the terms are all positive. Therefore, if we prove that the series on the right has the same terms as the one on the left, given that the latter converges, by Theorem A.1 the series on the right must also converge to the same sum, as established by the thesis of the Theorem.
First of all, we have to observe that, in the particular case d = 1, the product on the left does not contain terms (because 1 has no prime factors) and therefore by convention it’s equal to 1; on the right, however, the only a to consider is 1, because the only positive integer in the factorization of which only prime factors of 1 appear (i.e. without prime factors) is 1 itself; ultimately, we’ll obtain the equality 1 = \frac{1}{1} which is trivially true.
Now let’s examine the case d \gt 1, which assures us that there exists at least one prime divisor of d. Since d has no squares, we can set d := p_1 \cdot \ldots \cdot p_k, obtaining:
where in step (a) we implicitly applied Theorem A.1, while in step (b) we observed that the terms of the first summation are of the type \frac{1}{p_1^{i_1}}, with i_1 non-negative (for i_1 = 0 we obtain the first term, which is 1), and so on for the other summations, up to the k-th.
It remains to be proved that:
To prove this equality, again by Theorem A.1, we must prove that the two series have the same terms, that is, that all the elements of the left summation also appear in the right summation, and vice versa.
As for the left summation, it’s evident that in the number p_1 \cdot \ldots \cdot p_k there are only the prime factors of d, which are precisely p_1, \ldots, p_k, therefore this number will be one of the as of the right summation. Conversely, if a is a number such that its factorization contains only prime factors of d, i.e. only p_1, \ldots, p_k, it can be written in the form p_1^{i_1} \cdot \ldots \cdot p_k^{i_k}, with i_1, \ldots, i_k \geq 0 (where 0 allows us to express the absence of one or more prime numbers in the factorization); therefore we’ll find the term \frac{1}{a} in the left summation. The proof is thus complete.