A Simplified Explanation of What It Means to Assign a Finite Value to an Infinite Sum

Olga Kosheleva1 and Vladik Kreinovich1 Departments of 1Teacher Education and 2Computer Science University of Texas at El Paso, El Paso, TX 79968, USA [email protected], [email protected]

Abstract—Recently, a video made rounds that explained that II.HOW CANA DIVERGENT INFINITE HAVE A it often makes sense to assign finite values to infinite sums. For FINITE SUM:ASIMPLE EXAMPLE example, it makes sense to claim that the sum of all natural is equal to −1/12. This has picked up interested in An example. To explain how a divergent infinite series can media. However, judged by the viewers’ and readers’ comments, be assigned a finite sum, let us start with trying to define the for many viewers and readers, neither the video, not the cor- sum of another infinite series: responding articles seem to explain the meaning of the above 1 2 3 4 inequality clearly enough. One of the main stumbling blocks is the 1 + 2 + 2 + 2 + 2 + ... = fact that the infinite sum is clearly divergent, so a natural value of the infinite sum is infinity. What is the meaning of assigning a 1 + 2 + 4 + 8 + 16 + ... finite value to this (clearly infinite) sum? While the explanation of the above equality is difficult to describe in simple terms, the If we simply keep adding these numbers, we get larger and main idea behind this equality can be, in our opinion, explained larger values: rather naturally, and this is what we do in this paper. • we start with 1, • then, we get 1 + 2 = 3, I.INTRODUCTION • after that, we get (1 + 2) + 4 = 3 + 4 = 7, − The sum of all positive integers is equal to 1/12: a • statement. A recent video [8] explains that it often makes etc. sense to assign finite values to infinite sums. For example, it In the limit n → ∞, the sum 1+22 +...+2n tends to infinity. makes sense to claim that 1 1 + 2 + 3 + 4 + ... = − . 12 Question. In what sense can we then declare some finite to be the sum of the above infinite series? This has picked up interested in other media sources, including an article in the New York Times [7]. Main idea. To be able to explain how the sum can be finite, let us do the following: A clarification is needed. Judged by the viewers’ and readers’ • comments, for many viewers and readers, neither the video, not instead of considering just the infinite series itself, the article seem to explain the meaning of the above inequality • let us consider a family of infinite series – of which clearly enough. One of the main stumbling blocks is the fact includes the desired series is a particular case. that the infinite sum 1 + 2 + ... Let us apply this general idea to our simple example. For the infinite series is clearly divergent: as n increases, the sum 1 + 21 + 22 + 23 + 24 + ..., n · (n + 1) 1 + 2 + ... + n = a natural generalization is a family of geometric progression 2 series 2 3 tends to infinity, so a natural value of the infinite sum is infinity. 1 + q + q + q + ... What is the meaning of assigning a finite value to this corresponding to different values of the parameter q: (clearly infinite) sum? • For q = 2, we get the desired series. • What we do in this paper. While the explanation of the above For other values of q, we get different infinite series. equality is difficult to describe in simple terms, the main idea behind this equality can be, in our opinion, explained rather When |q| < 1, we can get an explicit expression for the naturally. This is what we do in this paper. sum. In particular, when |q| < 1, the sum of the geometry progression series has a well-defined value: the limit of the The resulting sum for the simple example with which we values started. In particular, for q = 2, we get 1 + q + ... + qn 1 − s(2) = − = 1. when n tends to infinity. 1 2 Thus, according to to the above definition, we have To find this limit, we can take into account that if we multiply the sum 1 + 21 + 22 + 23 + 24 + ... =

def n 1 + 2 + 4 + 8 + 16 + ... = −1. sn = 1 + q + ... + q by q, we get III.CAN WE USE THIS IDEATO COMPUTETHE SUMOF 2 n n+1 ALL NATURAL NUMBERS:FIRST ATTEMPT q · sn = q + q + ... + q + q . Idea. How can we apply the general idea to the computation · So, if we subtract q sn from sn, all the terms cancel out except of the sum of all natural numbers? for 1 and qn+1, and so we get The above geometric progression series lead to the follow- n+1 sn − q · sn = 1 − q , ing natural idea: let us consider the sum 1 2 n from which we can conclude that 1 · q + 2 · q + ... + n · q + ... 1 − qn+1 This series converges for |q| < 1, and for q = 1, we get exactly s = . n 1 − q the desired sum of all natural numbers. So maybe this is a way that we can get the desired sum? When n tends to infinity, then qn+1 tends to 0, so the Let us get an explicit expression for the sum when |q| < expression for s tends to n 1. Similarly to the case of the geometric progression, when 1 |q| < 1, the sum of the geometry progression series has a . 1 − q well-defined value: the limit of the values 2 n Thus, we get 1 · q + 2 · q + ... + n · q 1 when n tends to infinity. s(q) def= 1 + q + q2 + q3 + q4 + ... = . 1 − q To find this limit, we can take into account that if we multiply the sum For different values of the parameter q, we get different def 2 k n values of this sum. In other words, the sum s(q) is a function sn = 1 · q + 2 · q + ... + k · q + ... + n · q of q. by q, we get 2 3 k k+1 How can we extend this expression to the general case? q · sn = 1 · q + 2 · q + ... + (k − 1) · q + k · q + Strictly speaking, the left-hand side sum is defined only when − · n · n+1 |q| < 1, because otherwise, the series diverge. However, the ... + (n 1) q + n q . 1 k expression s(q) = is an analytical function of the So, if we subtract q · sn from sn, then, for each power q with 1 − q k ≤ n, we get the coefficient k − (k − 1) = 1, and we also parameter – in the sense that it can be expanded in the power get the term −n · qn+1: series. s − q · s = q + q2 + ... + qn − n · qn+1, By using these , we can naturally define the n n 1 from which we can conclude that values of this function s(q) = − for the values q for which 1 q 1 ( ) |q| ≥ 1. s = · q + q2 + ... + qn − n · qn+1 , n 1 − q The only value of the parameter q for which we cannot i.e., that produce any finite value for s(q) is is the value q = 1, because − · n+1 in this case, we have 1 q = 0 and thus, the expression q n−1 n q 1 sn = · (1 + q + ... + q ) − . s(q) = is infinite. 1 − q 1 − q 1 − q When n tends to infinity, then n · qn+1 tends to 0, and the Thus, for these q, we can define the sum of the correspond- geometric progression ing infinite series as the value of the function 1 + q + ... + qn−1 1 s(q) = ; 1 − q tends to 1 . see, e.g., [4]. 1 − q Thus, we get thus becomes the desired sum q s(q) def= 1 · q + 2 · q2 + 3 · q3 + 4 · q4 + ... = . 1 + 2 + ... (1 − q)2

For different values of the parameter q, we get different values of this sum. In other words, the sum s(q) is a function Estimating the sum for different values of s. The general of q. series converges for s > 1; the sum of this series is known as the ζ(s); see, e.g., [1], [3], [5], [6], We can extend this expression to the general case. Similarly [9], [10], [11]. to the case of the geometric progression, we can extend the above formula to other values of q. In particular, for q = 2, Comment. Computation of the values of this function is not we have as straightforward as the above computation of the sum of the 2 s(2) = = 2, geometric progression, but for some values s, the correspond- (1 − 2)2 ing result can be rather easily explained. As an example, let and thus, get yet another finite sum for the infinite series that us show how to compute the value ζ(2). tends to infinity: This computation is related to the fact that a polynomial · · 2 · 3 · 4 n n−1 1 2 + 2 2 + 3 2 + 4 2 + ... = 2. P (x) = x + an−1 · x + ... + a1 · x + a0

of n-th degree with n different roots x1, . . . , xn can be represented as the product This idea does not work for the sum of all natural numbers. There is only one value q for which this idea does not work, P (x) = (x − x1) · ... · (x − xn). q = 1 and this is exactly the case in which we are interested. This expression is tailored towards polynomials in which the So, to compute the sum of all natural numbers, we need coefficient at the highest degree is 1. We can easily come up to come up with another general family containing this sum. with the form tailored towards the polynomials in which the constant is 1, by dividing both sides of the above equality by · · IV. COMPUTINGTHE SUMOF ALL NATURAL NUMBERS: the product x1 ... xn. As a result, we get the expression SECOND ATTEMPTANDTHE RESULT 1 + a · x + a · x2 + ... + a · xn = ( 1 ) ( 2 ) (n ) How to apply the above idea to the sum of natural x x x numbers: need for an alternative expression. We want to 1 − · 1 − · ... · 1 − . x x x find a general family: 1 2 n It is reasonable to expect that for limits of polynomials – i.e., • that would contain the desired infinite series for analytical functions – we get a similar formula 1 + 2 + 3 + 4 + ... 1 + a · x + a · x2 + ... + a · xn + ... = ( 1 ) ( 2 ) (n ) as a particular case – and x x x 1 − · 1 − · ... · 1 − · ... • for which we would be able to extend the analytical x1 x2 xn expression to this case to get a finite answer. As an example, let us consider the function To get a , a natural idea is to raise things sin(x) . to the power, this will often guarantee convergence. In the x previous section, we have tried multiplying natural numbers by powers of some other number, this did not work out. A natural We know how to expand sin(x) into : idea is then to consider the powers of the natural numbers x3 x5 themselves. In other words, let us consider the following family sin(x) = x − + − ..., 3! 5! of infinite series: so 1 1 1 sin(x) x2 x4 1 + + + + ... = 1 − + − ... 2s 3s 4s x 3! 5! s We know all the zeros of this ratio function – they are the value For different values of the parameter , we get different  · infinite series. In particular, for s = −1, each term n π corresponding to positive natural numbers n. Thus, sin(x) 1 = s x n ( ) ( ) ( ) ( ) x x x x takes the form 1 − · 1 + · 1 − · 1 + · ... 1 1 π π 2 · π 2 · π − = = n, n 1 1/n For each pair of roots n · π, we get ( ) and the above sum ( ) ( ) 2 1 x x x 1 + + ... 1 − · 1 + = 1 − . 2s n · π n · π n2 · π2 Thus, Riemann’s hypothesis is probably the most well-known sin(x) = open problem in [1], [2], [3], [5], [6], [9], [10], x [11]. ( ) ( ) ( ) x2 x2 x2 s = −1 1 − · 1 − · ... · 1 − · ... In this paper, we are interested in the value . For π2 22 · π2 n2 · π this s, the resulting analytical function ζ(s) attains the value 1 When we multiply all these polynomials, we can take into ζ(−1) = − . account that the only terms proportional to x2 are terms 12 coming from the case when all multiplied coefficients are This is the meaning behind the above equality 1s except for one of them – else we would have a term 4 6 1 proportional to x or x , etc. In general, we have 1 + 2 + 3 + 4 + ... = − . 2 2 12 (1 + a1 · x ) · (1 + a2 · x ) · ... = 2 ACKNOWLEDGMENTS 1 + (a1 + a2 + ...) · x + ..., This work was supported in part by the National Science i.e., in our case, sin(x) Foundation grants HRD-0734825 and HRD-1242122 (Cyber- = ShARE Center of Excellence) and DUE-0926721. x ( ) 1 1 1 The authors are very thankful to Mourat Tchoshanov for x2 · + + ... + + ... + ... helping us clarify our text. π2 22 · π2 n2 · π2 On the other hand, we know that the coefficient at x2 is the REFERENCES Taylor expansion of the ration function is equal to [1] T. M. Apostol, “Zeta and related functions”, in F. W. Olver, 1 1 1 D. M. Lozier, R. F. Boisvert, and C. W. Clark, NIST Handbook − = − = − , of Mathematical Functions, Cambridge University Press, Cambridge, 3! 1 · 2 · 3 6 Massachusetts, 2010, pp. 601–616. thus, [2] P. Borwein, S. Choi, B. Rooney, and A. Weirathmueller, The Riemann 1 1 1 1 Hypothesis: A Resource for the Afficionado and Virtuoso Alike, Springer = + + ... + + ... 6 π2 22 · π2 n2 · π2 Verlag, New York, 2008. [3] H. M. Edwards, Riemann’s Zeta Function, Dover Publications, 2 Multiplying both sides of this equality by π , we conclude that New York, 1974. 2 [4] G. H. Hardy, , Clarendon Press, Oxford, UK, 1949. def 1 1 1 π ζ(2) = 1 + 2 + 2 + ... + 2 + ... = . [5] A. Ivic, The Riemann Zeta Function, John Wiley & Sons. New York, 2 3 n 6 1985.

4 6 [6] A. A. Karatsuba and S. M. Voronin, The Riemann Zeta-Function, By comparing coefficients at x , x , etc., we can similarly W. de Gruyter, Berlin, 1992. compute the values ζ(4), ζ(6), etc. [7] D. Overbye, “In the end, it all adds up to −1/12”, New York Times, February 3, 2014, http://www.nytimes.com/2014/02/04/science/in-the- Extending this function to different values s. The function end-it-all-adds-up-to.html ζ(s) is analytical, and it can be extended to values s ≤ 1. [8] T. Padilla, E. Copeland, and B. Haran, “1 + 2 + 3 + 4 + ... – feat”, http://www.numberphile.com/videos/analytical continuation1.html It is interesting to mention that this extension was first [9] B. Riemann, “Uber¨ die Anzahl der Primzahlen unter einer gegebe- discovered by B. Riemann in the 19th century [9]. Riemann nen Grosse”,¨ Monatsberichte der Berliner Akademie, November 1859; actually extended this function not just to real values s ≤ 1, reprinted in [2], [3], [10]. but also to complex values of s. Riemann showed that ζ(s) = 0 [10] B. Riemann, Gesammelte Werke, Teubner, Leipzig, 1892; reprinted by for s = −2, z = −4, z = −6, etc. He also noticed – but could Dover Publ., New York, 1953. not prove it – that all other zeros of this function have real part [11] E. C. Titchmarsh, The Theory of the Riemann Zeta Function, Oxford equal to 0.5. This hypothesis of Riemann leads to interesting University Press, Oxford, UK, 1986. consequences about prime numbers – but until now, no one has been able to prove or disprove this hypothesis.