On the sums of series of reciprocals

arXiv:math/0506415v1 [math.HO] 20 Jun 2005

On the sums of series of reciprocals

∗

Leonhard Euler†

1. So hard already have the study and investigation of series of reciprocals of the natural numbers been that a demonstration of anything new about them is seen to be able to be found only with difficulty. For whenever the summation of series is considered, with which in general the sum of each series of that type may be investigated, not any method is able to express them in this way. I have already many times, with various methods of summation that I have related, carefully dealt with these series, still neither have I obtained any others, having struggled so that I may have specified the sum of them nearly exactly or have reduced them with the utmost of my labors to a square of transcendentals; of those others that have been advanced in the preceding dissertation, this indeed betters the ones that have come before. I speak here in fact about series of fractions of which the numerators are 1 and indeed the denominators are the squares, the cubes, or other ranks, of 1 1 the natural numbers; of this model are 1 + 14 + 19 + 16 + 25 + etc., likewise 1 1 1 1 + 8 + 27 + 64 + etc., and similarly for higher powers, of which the general terms are contained in the form x1n . 2. I have recently deduced entirely unexpectedly an elegant expression 1 + etc., which rests on a circle which for the sum of this series 1 + 41 + 91 + 16 ∗

Delivered to the St.–Petersburg Academy December 5, 1735. Originally published as De summis serierum reciprocarum, Commentarii academiae scientiarum Petropolitanae 7 (1740) 123–134 and reprinted in Leonhard Euler, Opera Omnia, Series 1: Opera mathematica, Volume 14, Birkh¨ auser, 1992. A copy of the original text is available electronically at the Euler Archive, at http://www.eulerarchive.org. This paper is E41 in the Enestr¨om index. † Date of translation: June 17, 2005. Translated from the Latin by Jordan Bell, 3rd year undergraduate in Honours Mathematics, School of Mathematics and Statistics, Carleton University, Ottawa, Ontario, Canada. Email: [email protected]. This translation was written during an NSERC USRA supervised by Dr. B. Stevens.

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is to be squared, so that if the true sum of this series were obtained, from it at once the square of a circle would follow. I have discovered for the sum of six of this series to be equal to the square of the circumference of a circle whose diameter is 1; or if the sum of this series is put as equal to s, it will √ hold for the ratio of 6s to 1 to be as the circumference to the diameter. Moreover, I have recently shown for the sum of this series to be approximately 1,6449340668482264364, from which in truth six of this number, if the square root is removed, produce the number 3,141592653589793238, expressing the circumference of a circle whose diameter is 1. Then again in the footsteps by 1 1 1 1 which I uncovered this sum follows the series 1+ 16 + 81 + 256 + 625 + etc., whose sum I have likewise discovered to depend on the square of a circle. Namely, the sum of this multiplied by 90 gives the biquadrature of the circumference of a circle whose diameter is 1. Likewise, by similar reasoning I have been able to determine the sums of the following series, in which the exponents are of ranks of even numbers. 3. To this end therefore, with it most helpful for me to reveal the way in which I obtained this, I will explain in order the whole of the property which I made use of. In the circle AMBNA having been described with Figure 1 center C and radius AC, or also BC, equal to 1, I contemplated an arc AM whose sine is MP , and cosine indeed CP . It then having been put AM = s, with sine P M = y and cosine CP = x, by a method which is already well known, according to the arc s having been given, the sine y and also the cosine x are able to be defined by a series, namely, as may been seen here s3 s5 s7 and indeed everywhere, it is y = s − 1·2·3 + 1·2·3·4·5 − 1·2·3·4·5·6·7 + etc. and too s4 s6 s2 x = 1 − 1·2 + 1·2·3·4 − 1·2·3·4·5·6 + etc. In particular, from the consideration of these equations I arrived at the sums of the series related above, of which the second of the equations can be directed in nearly the same direction as the other, and on account of this circumstance, it will be possible to treat the other in as much the same way as I will set forth this. 3

5

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s s s 4. The leading equation y = s − 1·2·3 + 1·2·3·4·5 − 1·2·3·4·5·6·7 + etc. thus expresses the relation between an arc and its sine. By means of this, with the arc having been given the sine itself, and with the sine having been given the arc itself, will be able to be determined. I will have considered though the sine y as having been given, and I investigate how the arc s may be able to be elicited from y. At this point however, before anything else, it is to be observed for countless arcs to correspond to the same sine y,

2

M

m

P

A

p C

n

N

3

B

hence the given equation will be bound to be satisfied by countless arcs. If indeed in this equation s were seen as an unknown, which has infinitely many dimensions, it is not surprising that if then this equation contains countless simple factors, each of them having been set equal to nothing, it ought to give the appropriate value for s. 5. To the extent that the factors of this equation were to be known, likewise all the roots of it, that is the values of s itself, would be known, and thus in turn, if all the values of s itself will be able to be assigned, then likewise all the factors of it will be obtained. For the purpose that I might be able to judge as well about the roots as about the factors, I transform s3 s5 the given equation into this form: 0 = 1 − ys + 1·2·3·y − 1·2·3·4·5·y + etc. If now all the roots of this equation, that is all the arcs, the sine of which is the very same y, were A, B, C, D, E, etc., then too the factors will be the quantities 1 − As , 1 − Bs , 1 − Cs , 1 − Ds , etc. On account of this, it will be s3 s5 1 − ys + 1·2·3·y − 1·2·3·4·5·y + etc. = (1 − As )(1 − Bs )(1 − Cs )(1 − Ds ) etc. 6. From the nature and also from the resolving of equations, it is established for the coefficient of the term to which s belongs, that is y1 , to be equal to the sum of all the coefficients of s itself in the factors, that is 1 = A1 + B1 + C1 + D1 + etc. Next, the coefficient of s2 itself, which is equal to y 0, because this term is absent from the equation, is equal to the aggregate of factors two apiece from the terms in the sequence A1 , B1 , C1 , D1 , etc. In turn, 1 − 1·2·3·y will be equal to the aggregate of factors four apiece from the terms in the same sequence A1 , B1 , C1 , D1 , etc. In a similar way, it will be that 0 is equal to the aggregate of factors four apiece from the terms of the same sequence, 1 and that + 1·2·3·4·5·y is equal to the aggregate of factors five apiece from the terms of this sequence, and so on. 7. Having put moreover the smallest arc AM = A whose sine is P M = y, and the semicircumference of the circle equal to p, A, p − A, 2p + A, 3p − A, 4p + A, 5p − A, 6p + A, etc., together with −p − A, −2p + A, −3p − A, −4p + A, −5p − A, etc., will be all the arcs whose sine is the very same y. Thus, before we use the sequence A1 , B1 , C1 , D1 , it is changed into this: A1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , etc. Therefore the sum p−A −p−A 2p+A −2p+A 3p−A −3p−A 4p+A −4p+A of all these terms is equal to y1 ; on the other hand, the sum of the factors of this sequence two apiece is equal to 0; the sum of the factors three apiece is −1 ; the sum of the factors four apiece is equal to 0; the sum of equal to 1·2·3·y 4

+1 the factors five apiece is equal to 1·2·3·4·5·y ; the sum of the factors six apiece is equal to 0, and thus so further on.

8. Furthermore, if such a series is had: a + b + c + d + e + f + etc., whose sum were α, the sum of the factors two apiece were equal to β, the sum of the factors three apiece were equal to γ, the sum of the factors four apiece were equal to δ, etc., the sum of the squares of each of the terms will be a2 + b2 + c2 + d2 + etc. = α2 − 2β; truly too, the sum of the cubes a3 + b3 + c3 + d3 + etc. = α3 − 3αβ + 3γ; the sum of the biquadratures would be equal to α4 − 4α2 β + 4αγ + 2β 2 − 4δ. On the other hand, so that it is clearly apparent the way in which these formulas progress, we put the sum of the terms a, b, c, d, etc. to be equal to P , the sum of the squares to be equal to Q, the sum of the cubes to be equal to R, the sum of the biquadratures to be equal to S, the sum of the fifth powers to be equal to T , the sum of the sixths to be equal to V , etc. This having been set, it will be P = α; Q = P α − 2β; R = Qα − P β + 3γ; S = Rα − Qβ + P γ + 4δ; T = Sα − Rβ + Qγ − P δ + 5ǫ; etc. 1 9. Therefore in our case the sum of all the terms of the series A1 , p−A , 1 1 1 1 1 , , , , etc., or α, is equal to y ; the sum of the factors −p−A 2p−A −2p−A −3p−A −1 two apiece or β is equal to 0, and too for the higher γ = 1·2·3·y , δ = 0, +1 ǫ = 1·2·3·4·5·y , ζ = 0, etc., it will be for the sum of these terms P = y1 , the sum of the squares of these terms Q = Py = y12 , the sum of the cubes of 1 P , the sum of the biquadratures S = Ry − 1·2·3·y . these terms R = Qy − 1·2·y Q S 1 T R P Further likewise, T = y − 1·2·3·y + 1·2·3·4·y , V = y − 1·2·3·y + 1·2·3·4·5·y , W = Q V S 1 − 1·2·3·y + 1·2·3·4·5·y − 1·2·3·4·5·6·y . By this route, a rule for the sums of higher y powers can easily be determined.

10. We put now the sine P M = y equal to the radius, so that it is y = 1, the smallest arc A whose sine is 1 will be a quarter part of the circumference, equal to 21 p, or by denoting with q a quarter part of the circumference, it will be A = q and p = 2q. Therefore the last series will 1 be changed into 1q , 1q , −1 , − 3q1 , +1 , + 5q1 , − 7q1 , − 7q1 , + 9q , + 9q1 , etc., where 3q 5q the terms come forth as equal in pairs. The sum of these terms therefore, 1 which is 2q (1 − 13 + 51 − 17 + 91 − 11 + etc.) is equal to P = 1. Hence it is then 1 1 1 1 1 seen 1 − 3 + 5 − 7 + 9 − 11 + etc. = q2 = p4 . Thus this series quadrupled is equated to the semicircumference of the circle whose radius is 1, or the 5

entire circumference of the circle whose diameter is 1. Indeed, this is the very series already discovered by Leibniz, which specifies the square of a circle. Therefore because of this, if it by chance has not been seen to be adequately settled, support will shine forth for the greatness of this series; thus in this way for the remaining, which are derived by this method, nothing at all will be able to be doubted. 11. We now sum the squares of the terms that have been come upon for the case in which y = 1, which produces the series + q12 + q12 + 9q12 + 9q12 + 1 1 1 + 25q1 2 + etc., whose sum is q22 ( 11 + 91 + 25 + 49 + etc.), which is therefore 25q 2 bound to be equal to Q = P = 1. From this it follows for the sum of the 2 2 1 1 series 1 + 91 + 25 + 49 + etc. to be equal to q2 = p8 , with p denoting the total circumference of a circle whose diameter is equal to 1. On the other hand, 1 the sum of the series 1 + 91 + 25 + etc. is derived from the sum of the series 1 1 1 1 1 + 4 + 9 + 16 + 25 + etc., which having been lessened by a quarter part gives it. Therefore the sum of this series is equal to the sum of the previous series combined together with a third part of itself. On account of this, it will be 2 1 1 1 + 25 + 36 + etc. = p6 , and so the sum of this series multiplied 1 + 14 + 91 + 16 by 6 is equal to the square of the circumference of a circle whose diameter is 1, which is the very proposition which I had initially made mention of. 12. Therefore in the case in which y = 1 with it P = 1 and Q = 1, the other letters will be R, S, T, V , etc. such that it follows: R = 12 ; S = 13 ; T = 5 2 61 17 ; V = 15 ; W = 720 ; X = 335 , etc. Moreover, were the sum of the cubes 24 1 2 equal to R = 2 , it will be q3 (1 − 313 + 513 − 713 + 913 − etc.) = 12 . In turn it will 3

3

be 1 − 313 + 513 − 713 + f rac193 − etc.) = q4 = p32 , for which reason the sum of this series multiplied by 32 gives the cube of the circumference of a circle whose diameter is 1. In a similar way, the sum of the biquadratures, which 4 is p24 (1 + 3+ 514 + 714 + 914 + etc.) is bound to be equal to 18 , and then it will be 4

4

p 1 + 314 + 514 + 714 + 914 + etc.) = q6 = 90 . In fact, this series multiplied by 16 15 p4 ; is equal to 1 + 214 + 314 + 514 + 614 + etc., from which this series is equal to 90 1 1 1 that is, the sum of the series 1 + 24 + 34 + 44 + etc. multiplied by 90 gives the biquadrature of the circumference of a circle whose diameter is 1.

13. In a similar way the sums of higher powers may be discovered; it will 5 5p5 ; and come about that it is found 1 − 315 + 515 − 715 + 915 − etc. = 5q48 = 1536 q6 p6 1 1 1 1 then 1 + 36 + 56 + 76 + 96 + etc. = 15 = 960 . Indeed, with the sum of this series having been uncovered, at once the sum will be found of the series 6

6

p 1 + 216 + 316 + 416 + 516 + etc., which will be equal to 945 . Then for the seventh 61q 7 61p7 1 1 1 1 , and for the powers it will be 1 − 37 + 57 − 77 + 97 − etc. = 1440 = 184320 17q 8 17p8 1 1 1 1 eight 1 + 38 + 58 + 78 + 98 + etc. = 630 = 161280 ; from this it is deduced that p8 . Moreover, it is to be observed for the odd 1+ 218 + 318 + 418 + 518 + 618 + etc. = 9450 exponents of the powers of these series to alternate signs, while indeed those for the even powers are to be equal; from this it is derived that the sum of the general series 1 + 21n + 31n + 41n + etc. will be able to be exhibited, in the cases in which n is an even number. In addition, it is also to be noted that if the 5 2 61 17 general term were able to be assigned of the sequence 1, 1, 12 , 13 , 24 , 15 , 720 , 315 , etc., whose values we have found as the letters P, Q, R, S, etc., then from this the squaring of a circle will be exhibited.

14. In this we have put the sine P M equal to the radius, and we may therefore see what kind of series is produced if y itself takes other values. Therefore it is y = √12 , for which sine the smallest corresponding arc is 41 p. It having been then put A = 14 p, the series of single terms, that is of first powers, 4 4 4 4 4 will be p4 + 3p − 5p − 7p + 9p + 11p − etc., the sum P of which series is equal to √ 1 1 1 1 = 2. It will therefore be had that 2√p 2 = 1 + 31 − 15 − 17 − 19 − 11 − 13 − 15 y etc., for which series, the rule for the signs was discussed by Leibniz, and was advanced formerly by Newton. Indeed, the sum of the squares of these 1 1 (1 + 91 + 25 + 49 + textrmetc.) will be equal to Q = 2. Therefore it terms 16 p2 will be 1 + 91 +

1 25

+

1 49

+ etc. =

p2 , 8

just as had been found before.

√

3. [sic] If it is made y = 23 , the smallest arc corresponding to this sine will be 60 ◦, and therefore A = 13 p. From this case therefore, the following 3 3 3 3 3 series of terms 3p + 2p − 4p − 5p + 7p + 8p etc. is elicited, the sum of the 1 2 terms of which is equal to y = √3 itself. It will therefore be obtained that 2p √ = 1+ 1 − 1 − 1 + 1 − 1 − 1 − 1 + etc. Truly, the sum of the squares of these 3 3 2 4 5 7 8 10 11 2

1 1 1 1 = 1+ 14 + 16 + 25 + 49 + 64 + terms is equal to y12 = 34 ; from this it follows to be 4p 27 etc. in which series certain terms are absent. In fact, this is extracted from 2 1 the series 1 + 14 + 91 + 16 etc., whose sum was found to be equal to p6 ; on the other hand, if this series is made lessened by a ninth part, the very series 2 above is produced, whose sum therefore should be equal to p6 (1 − 19 ) = 4pp . 27 In a similar way, if another sine is assumed, another series will be produced, whose simple terms are squares of higher powers, of which the sums involve the square of a circle.

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16. However, if it is put y = 0, no other series will be able to be assigned in this manner, because y is situated in the denominator, that is, the initial equation is divided by y. But in another way series will be able to be deduced from this, which will be together with the series 1 + 21n + 31n + 41n + etc. if n is an even number: to the extent that the sums of these series are to have been discovered, I will settle separately the case in which y = 0. Indeed, it having been put y = 0, from which the fundamental equation will turn into s3 s5 57 this 0 = s − 1·2·3 + 1·2·3·4·5 − 1·2·3·4·5·6·7 + etc. The roots of this equation give all the arcs of which the sine est equal to 0. Moreover, the single smallest root is s = 0, whereby the equation divided by s will exhibit all the remaining arcs of which the sine is equal to 0; these arcs will hence be the roots of this equation s2 s4 s6 0 = 1 − 1·2·3 + 1·2·3·4·5 − 1·2·3·4·5·6·7 + etc. Truly then, those arcs of which the sine is equal to 0 are p, −p, +2p, −2p, 3p, −3p, etc., of which the the second of the two of each pair is negative, each of these because the equation indicates for the dimensions of s to be even. Hence the divisors of this equation will be s s , 1+ 2p , etc. and by the joining of these divisors two by two it 1− ps , 1+ p1 , 1− 2p 2

4

6

2

2

2

2

s s s s s s will be 1− 1·2·3 + 1·2·3·4·5 − 1·2·3·4·5·6·7 + etc. = (1− ps2 )(1− 4p 2 )(1− 9p2 )(1− 16p2 ) etc.

17. It is now clear from the nature of equations for the coefficient of ss 1 1 that is 1·2·3 to be equal to p12 + 4p12 + 9p12 + 16p 2 + etc. In fact, the sum of the 1 , and factors from the terms two apiece of this series will be equal to 1·2·3·4·5 1 the sum of the factors by three equal to 1·2·3·4·5·6·7 etc. On account of this 1 1 1 fact, it will be like §8, α = 1·2·3 ; β = 1·2·3·4·5 ; γ = 1·2·3·4·5·6·7 , etc., and likewise, 1 1 1 1 the sum of the terms having been put p2 + 4p2 + 9p2 + 16p2 + etc. = P , the sum of the squares of these terms equal to Q, the sum of the cubes equal to R, the sum of the biquadratures equal to S, etc.; by means of §8 it will 1 1 1 be P = α = 1·2·3 = 61 ; Q = P α − 2β = 90 ; R = Qα − P β + 3γ = 945 ;S = 1 1 Rα − Qβ + P γ − 4δ = 9450 ; T = Sα − Rβ + Qγ − P δ + 5ǫ = 93555 ; V = 691 T α − Sβ + Rγ − Qδ + P ǫ − 6ζ = 6825·93555 , etc. 18. From these therefore, the following sums are derived:

8

1+ 1+ 1+ 1+ 1+ 1+

1 + 312 + 412 + 512 22 1 + 314 + 414 + 514 24 1 + 316 + 416 + 516 26 1 + 318 + 418 + 518 28 1 + 3110 + 4110 + 210 1 + 3112 + 4112 + 212

2

etc. = p6 = P p4 etc. = 90 =Q p6 etc. = 945 = R p8 etc. = 9450 =S p10 1 etc. = 93555 = T 510 691p12 1 etc. = 6825·93555 = V. 512

Those series for the higher powers, although with much labor, are able to be produced from the given rule. Moreover, by dividing each each of these series by the preceding ones, the following equations arise: p2 = 6P = 15Q = P 10S 99T 6825V 21R = = = , etc., from which each of the expressions is equated 2Q R 10S 691T with the square of the circumference whose diameter is 1. 19. However, even if the approximate sums of these series are able to be easily exhibited, still to be sure they are not able to offer much of help about the circumference of a circle, because of the root having been squared, which ought to be extracted; from the first series we will draw forth these expressions which are equal to the circumference p itself. It will advance such that it follows:   1 1 1 1 1 p = 4 1 − 3 + 5 − 7 + 9 − 11 + etc.  1 1 1 1 1  1+ 2 + 2 + 2 + 2 + 2 + etc. 3 5 7 9 11 p=2 1− 1 + 1 − 1 + 1 − 1 + etc.  1 3 1 5 71 9 1 11 1  1− 3 + 3 − 3 + 3 − 3 + etc. 3 5 7 9 11 p = 4 1+ 1 + 1 + 1 + 1 + 1 + etc.   312 512 712 912 1112 1+ 4 + 4 + 4 + 4 + 4 + etc. 3 5 7 9 11 p = 3 1− 1 + 1 − 1 + 1 − 1 + etc.  331 531 731 931 1131  16 1− 35 + 55 − 75 + 95 − 115 + etc. p = 5 1+ 1 + 1 + 1 + 1 + 1 + etc.  314 514 714 914 1114  25 1+ 36 + 56 + 76 + 96 + 116 + etc. p = 8 1− 1 + 1 − 1 + 1 − 1 + etc.  351 551 751 951 1151  192 1− 37 + 57 − 77 + 97 − 117 + etc. p = 61 1+ 1 + 1 + 1 + 1 + 1 + etc. 36

56

76

9

96

116