Oservations on armoni Progressions * Leonard Euler Under te name of armoni progressions all series of frations are understood, wose numerators are equal to ea oter, ut wose denominators on te oter onstitute an aritmeti progression. Terefore, a general form of tis kind is a, a +, For, ea tree ontiguous terms, as a +, +, a +, a + 3 a + 3, ave tis property tat te differenes of te outer ones from te middle term are proportional to te outer terms temselves. Of ourse, it is a + a + : a + a + 3 = et. a + : a + 3. But eause tis is te property of te armoni proportion, series of frations of tis kind were alled armoni progressions. Te ould also ave een alled reiproals of first order, sine in te general term te index n as a+(n ) one, more preisely one negative, dimension. *Original title: De Progressionius armoniis Oersavatioes, first pulised in Commentarii aademiae sientiarum Petropolitanae 7 (734/35), 740, p. 50-6, reprinted in in Opera Omnia: Series, Volume 4, pp. 87-00, Eneström-Numer E43, translated y: Alexer Ayok for Euler-Kreis Mainz
Altoug in tese series te terms ontinuously derease, te sum of a series of tis kind ontinued to infinity is neverteless always infinite. To demonstrate tis no metod to sum tese series is neessary, ut te validity will easily eome lear from te following priniple. A series, wi ontinued to infinity as a finite sum, even if it is extended twie as far, will otain no augmentation, ut tat, wat is added after infinity y ogitation, will in reality e infinitely small. For, if tis would not e te ase, te sum of series, even if ontinued to infinity, would not e determined terefore not finite. From tis it follows, if tat, wat arises from te ontinuation eyond te infinitesimal term, is of finite magnitude, te sum of te series neessarily must e infinite. Terefore, from tis priniple we will e ale to deide, weter te sum of a given series is infinite or finite. 3 Terefore, let te series a, a +, a + e ontinued to infinity te infinitesimal term, wile i is an a+(i ) infinite numer, wi is te index of tis term. Now, ontinue tis series furter from te term until te term, wose exponent is ni. a+i et. a+(ni ) Terefore, te numer of additionally added terms is (n )i. But teir sum will e smaller tan ut larger tan (n )i a + i, (n )i a + (ni ). But eause i is infinitely large, a vanises in ea of ot denominators. Hene te sum will e greater tan ut smaller tan (n ), n (n ).
From tis it is perspiuous tat tis sum is finite as a logial onsequene te sum of te propounded series a, a+ ontinued to infinity is infinitely large. 4 But loser limits of tis sum of te terms from i to ni are found from te following properties of te armoni proportion. Of ourse, every armoni proportion is of su nature tat te middle term is smaller tan te tird part a+i of te sum of all tree. Terefore, te middle term etween wi is, multiplied y te numer of terms (n i)i or a+ ni+i (n )i a + ni+i a+(ni ), will e smaller tan te sum of te terms. Or te sum of te terms ene will e greater tan (n ) (n + ) eause of te infinite i. Furtermore, te aritmeti mean of te most outer terms is greater tan te tird part of te sum of te terms. From tis it follows tat also in te armoni series te sum of terms will e smaller tan (n )i times te aritmeti mean of te most outer terms, wi is (a + (ni + i )) (a + i)(a + (ni )) Hene te sum will e smaller tan or (n + ). ni su tat tese two limits are n ), n (n ) (n + ) ene te sum approximately (n ) n = (n ) n, wi is te proportional middle etween te limits. 3
5 From tese tings is it possile to onlude in wi ases tis more universal series a, a +, a + α et. to infinity until a + i α as a finite or infinite sum. For, let (n )i terms follow after te last term te sum of tese will e smaller tan ut greater tan (n ) i α, (n ) n α i α. Hene, if α was a numer greater tan unity, te sum of tese following terms will e = 0 terefore te sum of te progression will e finite. But if it is α <, te sum of te following terms will e infinite, wi is wy te sum of te progression itself will e infinite of an infinitely larger degree. Terefore, among tese progressions only te armoni, in wi it is α =, as tis property tat te sum of it ontinued to infinity is infinitely large ut te sum of te following terms after te infinitesimal term on te oter is finite. 6 But ow large te sum of terms from te term of te index i to te term of te index ni is, I investigate in te following way. Put te sum of te series a, a +,, a + (i ) until te term of te index i = s, wi is a quantity to e determined from a,, i. Let i grow y te unity s will ave te following term a+i as augmentation. Hene it will e Hene one finds di : ds = : a + i or ds = di a + i. s = C + log(a + i), wile C denotes a ertain onstant quantity. But it is also lear from tis form tat te sum of te same series ontinued from te eginning to te term of te index ni will e 4
= C + log(a + ni). Terefore, te differene of tese sums a + ni log a + i = log n will give te sum of te terms from limits of tis sum aove, (n ) n, or log > a+i to (wile a vanises) a+ni. But eause we assigned te log n will e greater tan (n ) (n+) (n ) n + log n < n n. smaller tan 7 Below we will sow tat tat quantity C is finite we will try to determine it. Terefore, C will vanis in te sum te sum of te progression a, a +, a + (i ). wile te numer of times is infinite = i, will eome = log(a + i) = log i. Terefore, te sum will e as te logaritm of te numer of terms ene infinitely smaller tan te root of aritrary large power of te numer of terms; neverteless it is infinitely large. 8 From tis onsideration innumerale series arise to denote te logaritms of ertain numers. At first, let us take tis armoni progression + + 3 + 4 + et., for wi it is a =, =, =. Terefore, te differene etween tis series + + 3 + + i ontinued to te term of te index i te same + + 3 + + ni 5
ontinued to te term of te index ni will e = log n. ene tat series sutrated from tis one leaves log n. But sine te numer of terms of tis series is n times greater tan te numer of te latter, from n term of te series + + + ni one as to sutrat one of te oter series + + + i, tat te sutration to infinity an e done in te same way. Hene it will e log n = + + + n + n + + + n + n + + + 3n + et. 3 Terefore, if te single terms of te inferior series are atually sutrated from te terms written over tem of te superior series for n te integer numer, 3, 4 et. are written, we will suessively otain te following series of logaritms: log = + 3 4 + 5 6 + 7 8 + 9 0 + + + et., log 3 = + 3 + 4 + 5 6 + 7 + 8 9 0 + + et., log 4 = + + 3 3 4 + 5 + 6 + 7 3 8 + 9 + 0 + 3 + et., log 5 = + + 3 + 4 4 5 + 6 + 7 + 8 + 9 4 0 + + + et., log 6 = + + 3 + 4 + 5 5 6 + 7 + 8 + 9 + 0 + 5 + et. et., Hene for te logaritm of ea numer a onvergent series is easily found. 6
9 From tese series oters of te same form, wi ave a rational sum, an e derived. For, sine te doule of te series = log is log 4, if te series is sutrated from tis one + + 3 3 4 + et. te remainder, namely, tis series + 3 4 + et., will e = 0, or 3 + 3 + 4 + 5 3 6 + et., = 3 + 4 + 5 3 6 + 7 + 8 + 9 3 0 + et. Similarly, if te series exiiting log 6 is sutrated from te sum of te series exiiting log log 3, te residue, namely will e = 0, or 3 4 + 5 + 6 + 7 8 9 0 + et., = + 3 + 4 5 6 7 + 8 + 9 + 0 et. In similar manner, one will e ale to find innumerale oter series of tis kind. 0 Tose series expressing te logaritms ertainly onverge, ut very slowly, wene, tat y means of tem te logaritms an onveniently e found, a ertain auxiliary tool is required. To find tis it must e noted tat tese series do not proeed uniformly, ut ave ertain revolutions, wi are asolved in so many terms as n as unities; terefore, I will all tat many terms taken simultaneously one memer of te series. So in te series for log two terms will onstitute one memer, in te series for log 3 tree, in te series for log 4 so fort. Terefore, te memers will onstitute an equal series to find logaritms it is neessary to add several memers. For, let us put tat m memers were added to find te logaritm of two instead of 7
all te following ones one will e ale to add 4m, wi will ome te loser to te trut, te greater te numer m was. To find log 3 to m already added memers instead of all te following ones add 9m. In similar manner for log 4 one must add 6m so fort. Tese remarks follow from te metod of summing applied in 6; sine in tis m must e a very large quantity, I negleted te numers added to m in te differential, tat te integration does not depend on logaritms. But to determine te sum, even toug it is infinite, of te series + + 3 + 4 + + i aurately, I express te single terms in te following way. It is = log + 3 3 + 4 5 + 6 7 + et. = log 3 + 4 3 8 + 4 6 5 3 + et., 3 = log 4 3 + 9 3 7 + 4 8 5 43 + et., 4 = log 5 4 + 6 3 64 + 4 56 5 04 + et.. i = log i + + i i 3 i 3 + 4 i 4 5 i 5 + et. Having added tese series it will arise + + 3 + + i = log(i + ) + 3 + 4 ( + 4 + 9 + ( + 8 + ) 6 + et. ) 7 + 64 + et. ) et. ( + 6 + 8 + 56 + et. Sine tese series are onvergent, if tey are summed approximately, it will arise 8
+ + 3 + + = log(i + ) + 0.5778. i If te sum is alled s, it will e, as we did it aove, ds = di i + ene s == log(i + ) + C. Terefore, we deteted te value of tis onstant C, wi is C = 0.5778. If te series + + 3 + + i is ontinued furter to infinity sudivided into memers, of wi ea as te series itself ontains i terms, te memer ontained witin i i will e = log, te following = log 3, te tird = log 4 3 et. And eause te series of te sum itself is te logaritm of infinity, one an analogously put log 0. And tis way we will otain te following rater urious seme: Series Sums + + + i log 0 + + i log + + 3i log 3 + + 4i log 4 3 + + 5i log 5 4 et. 3 It migt ertainly seem to e diffiult to find tese same properties of armoni logaritmi expressions analytially in te same way I used elsewere to sum series. But to anyone onsidering te sujet wit more attention it will eome lear tat tis annot only e done ut an even e done in mu more generality. For, I onsider not te simple armoni progression ut te one onneted a geometri progression, of wi kind tis one is x a + x a + + x3 a + + x4 a + 3 + et. I put its sum s aving multiplied ot y x a x a s = a x a + a+ x And aving taken differential one will ave a+ it will e a + + x a + + et. 9
( ) D.x a s = dx x a + x a a+ + x + et. = Having taken integrals again it will e a x dx x. x a s = a x dx x s = x a From tis series I now sutrat tis one a x dx x. f x m wose sum sall e t. Multiply it y g + f xm g + + f x3m g + + et., it will e m x m(g ) ; m x m(g ) t = mg f x mg + f x m(g+) And aving taken differentials it will e m(g+) m(g + ) + f x m(g + ) + et. m ( m(g ) D.x t = dx f x mg + f x m(g+) + f x m(g+) + et. Hene one will ave ) = f x mg dx x m. And ene t = f m x m(g ) mg x dx x m. s t = x a a x a dx x f m x m(g ) mg x dx x m. 0
But tis sutration as to e done in su a way tat from te term of te index m of te series s te first term of te series t is sutrated from te term of te index m of tat series te seond of tis series so fort. 4 To find our logaritmi series, let it e a = g =. Having done tis it will e s = dx x = log x Terefore t = f mx m dx x m = f log x m. s t = log ( xm ) f ( x) But tat tis expression eomes finite for x =, it must e f = ; terefore, let all tis letters eome = it will e s t = log xm x = log( + x + x + + x m ). Tis expression gives te differene etween tese series x + x + x3 3 + x4 4 + x5 5 + et. xm + xm + x3m 3 + et. Hene, if it is m =, it will e log( + x) = x x + x3 3 x4 4 + et.; if it is m = 3, it will e log( + x + x ) = x + x x3 3 + x4 4 + x5 5 x6 6 + et. in similar manner log( + x + x + x 3 ) = x + x + x3 3 3x4 4 + et..
If in tese it is x =, te same series for te logaritms of natural numers as tose we gave efore [ 8] will arise. 5 If it is = g, it will e Put x m = y; it will e t = f y t = f x m If furtermore it is a =, it will e mx m dx x m. dy ( y) y = f y log + y y = f x m log + x m. x m But s is te sum of tis series s = log x. gives tis series x a + x a + x3 3 + et. tx m = f log + x m x m f x m g Let a = g = ; it will e + f x 3m 3g + f x 5m 5g + et. s tx m = log x f log + x x m m = log ( ) f x m ( x) ( + x m ). f Tat tis expression eomes finite, if it is x =, it is neessary tat it is f = or f =. Terefore, let = m = n; te differenes of te series
x + x + x3 3 + x4 4 + et. will e x n + x3n 3 + x5n 5 + et. = log x n ( x)( + x n ). Put n = ; te differene will e = log +x +x for x = it will e = 0, wene tis series 3 + 3 + 4 + 5 3 6 + 7 + et. it will e = 0, as we already found aove [ 9]. 6 One an now find infinitely many oter series of tis kind aving a rational sum from tis form itself log +x +xx y assuming oter similar forms, wi vanis for x =. For, from tis form log +x, if it is expressed y +x means of a series, te found series itself immediately results. For, it is log( + x) = x x + x3 3 x4 4 + x5 5 et. log( + x ) = x x4 + x6 3 x8 4 + x0 5 et. Terefore, tis series sutrated from te superior one leaves tis one eind x 3x + x3 3 + x4 4 + x5 5 3x6 6 + et., wose sum will e log +x +x. In similar manner, log +x +x 3 will give tis series x x x3 3 x4 4 + x5 5 + x6 6 + x7 7 x8 8 x9 9 et. Terefore, aving put x = it will e 3
0 = 3 4 + 5 + 6 + 7 8 9 et., wi same we found already in 9. 7 In tis way one will e ale to find te sums of all irregular series of tis kind wi neverteless proeed regularly in memers; for, tey are always to e onsidered as te differene of two series. Let, e.g., tis series e propounded Tis is te differene of tese series + 3 + 4 5 + 6 + 7 8 + et. x + x + x3 3 + x4 4 + x5 5 + et. 3x + 3x5 5 + 3x8 8 + et. for x =. But te of tat series is log x, te sum of te first on te oter is 3xdx or x 3 log x + 3 log(x + x + ) + log x + 3 3 x + + 3 log 3 + 3. Terefore, aving sutrated tis one from tat one aving put x = log 3 + 3 log 3 + 3 3 3 3 log + 3 3 3 log 3+ 3 3 is 3 will arise for te sum of te propounded progression. But indeed te irumferene of te irle divided y 3 aving put te diameter = 3 log 3 + is its alf. 3 4
8 But even if te memers temselves go in non uniformly, te sum is assigned more diffiultly. Let us take tis series + 3 + 4 5 + 6 + 7 + 8 3 9 + 0 + + + 3 4 + et. Tis is te differene etween tese series + + 3 + 4 + 5 + + i i+3 + 3 5 + 4 9 + 5 4 + + i + i i+3 ontinued to infinity in su a way tat te most outer terms ave te same denominator i i+3. But te sum of te first of tese series is C + log i + log(i + 3) log, were C denotes te onstant found in, namely 0.5778. Te oter series wi is to e sutrated is resolved into tese two ( + 3 + 3 + 4 + + ) i ( 4 3 4 + 5 + 6 + + ) i + 3 Te sum of tat one 3 C + 3 log i, te sum of te first on te oter is 4 3 C 9 + 4 3 log(i + 3). Tese two sutrated from tat sum C + log i + log(i + 3) log leaves + 9 log or approximately.74078 for te sum of te propounded series. 5