Algebra

Summation of Series by the Method of Differences

Rule: From successive series of differences until a series of equal differences is obtained. Let 𝑎, 𝑏, 𝑐, 𝑑, ⋯ be the first terms of the several series; 264 then the 𝑛^th term of the given series is 𝑎+(𝑛−1)𝑏+(𝑛−1)(𝑛−2)1⋅2𝑐+(𝑛−1)(𝑛−2)(𝑛−3)1⋅2⋅3𝑑+265 The sum of 𝑛 terms =𝑛𝑎+𝑛(𝑛−1)1⋅2𝑏+𝑛(𝑛−1)(𝑛−2)1⋅2⋅3𝑐+⋯ Proved by Induction266

Example

𝑎⋯1+5+15+35+70+126+⋯
        𝑏⋯4+10+20+35+56+⋯
        𝑐⋯6+10+15+21+⋯
        𝑑⋯4+5+6+⋯
        𝑒⋯1+1+⋯

The 100^th term of the first series =1+99⋅4+99⋅981⋅26+99⋅98⋅971⋅2⋅34+99⋅98⋅97⋅961⋅2⋅3⋅4 The sum of 100 terms =100+100⋅991⋅24+100⋅99⋅981⋅2⋅36+100⋅99⋅98⋅971⋅2⋅3⋅44+100⋅99⋅98⋅97⋅961⋅2⋅3⋅4⋅5266 To interpolate a term between two terms of a series by the method of differences.

Example

Given ~~log~~ 71, ~~log~~ 72, ~~log~~ 73, ~~log~~ 74, it is required to find ~~log~~ 72⋅54. Form the series of differences from the given logarithms, as in (266).

 log 71log 72log 73log 74
            𝑎⋯1.85125831.85733251.86332291.8692317
            𝑏⋯.0060742.0059904.0059088
            𝑐⋯−.0000838−.0000816
            𝑑⋯−.000022
            considered to vanish.

~~log~~ 72⋅54 must be regarded as an interpolated term, the number of its place being 2⋅54. Therefore put 2⋅54 for 𝑛 in formula (265). Result ~~log~~ 72⋅54=1⋅8605777 267

Direct Factorial Series

Example

Ex 5⋅7⋅9+7⋅9⋅11+9⋅11⋅13+11⋅13⋅15+⋯ 𝑑=common difference of factors 𝑚=number of factors in each term 𝑛=number of terms 𝑎=first factor of first term −𝑑 𝑛^th term=(𝑎+𝑛𝑑)(𝑎+~~𝑛+1~~𝑑)⋯(𝑎+~~𝑛+𝑚−1~~𝑑)268

To find the sum of 𝑛 terms

Rule: From the last term with the next highest factor take the first term with the next lowest factor, and divide by (𝑚+1)𝑑. Proof by Induction. Thus the sum of 4 terms of the above series will be, putting 𝑑=2, 𝑚==3, 𝑛=4, 𝑎==3, 𝑆=11⋅13⋅15⋅17−3⋅5⋅7⋅9(3+1)2 Proved either by Induction, or by the method of Indeterminate Coefficients. 269

Inverse Factorial Series

Example

Ex. 15⋅7⋅9+17⋅9⋅11+19⋅11⋅13+111⋅13⋅15+⋯ Defining 𝑑, 𝑚, 𝑛, 𝑎 as before, the 𝑛^th term=1(𝑎+𝑛𝑑)(𝑎+~~𝑛+1~~𝑑)⋯(𝑎+~~𝑛+𝑚−1~~𝑑) 270 To find the sum of 𝑛 terms. Rule.: From the first term wanting its last factor take the last term wanting its first factor, and divide by (𝑚−1)𝑑. Thus the sum of 4 terms of the above series will be, putting 𝑑=2, 𝑚=3, 𝑛=4, 𝑎=3, 15⋅7−113⋅15(3−1)2 Proof: By Induction, or by decomposing theterms, as in the following example. 271

Example

Ex. To sum the same series by decomposing the terms into partial fractions. Take the general term in the simple form 2(𝑟−2)𝑟(𝑟+2) Resolve this into the three fractions 18(𝑟−2)−14𝑟+18(𝑟+2) by (235) Substitute 7, 9, 11, ⋯ successively for 𝑟, and the given series has for its equivalent the three series

18{ 15+17+19+111+113+⋯+12𝑛+3}
            +18{−27−29−211−213−⋯−12𝑛+3−12𝑛+5}
            +18{ 19+111+113+⋯+12𝑛+3+12𝑛+5+12𝑛+7}

and the sum of 𝑛 terms is seen, by inspection, to be 18{15−17−12𝑛+5+12𝑛+7}=14{15⋅7−1(2𝑛+5)(2𝑛+7)} a result obtained at once by the rule in (271), taking 15⋅7⋅9 for the first term, and 1(2𝑛+3)(2𝑛+5)(2𝑛+7) for the 𝑛^th or last term.272 Analogous series may be reduced to the types in (268) and (270), or else the terms may be decomposed in the manner shewn in (272).

Example

Ex: 11⋅2⋅3+42⋅3⋅4+73⋅4⋅5+104⋅5⋅6+⋯ has for its general term 3𝑛−2𝑛(𝑛+1)(𝑛+2)=−1𝑛+5𝑛+1−4𝑛+2 by (235) and we may proceed as in (272) to find the sum of 𝑛 terms. The method of (272) includes the method known as "Summation by Subtraction." The method of (272) includes the method known as "Summation by Subtraction", but it has the advantage of being more general and easier of application to complex series. 273

Composite Factorial Series

If the two series (1−𝑥)⁻⁵=1+5𝑥+5⋅61⋅2𝑥²+5⋅6⋅71⋅2⋅3𝑥³+5⋅6⋅7⋅81⋅2⋅3⋅4𝑥⁴+⋯ (1−𝑥)⁻³=1+3𝑥+3⋅41⋅2𝑥²+3⋅4⋅51⋅2⋅3𝑥³+3⋅4⋅5⋅61⋅2⋅3⋅4𝑥⁴+⋯ be multiplied together, and the coefficient of 𝑥⁴ in the product be equated to the coefficient of 𝑥⁴ in the expansion of (1−𝑥)⁻⁸, we obtain as the result the sum of the composite series 5⋅6⋅7⋅8×1⋅2+4⋅5⋅6⋅7×2⋅3+3⋅4⋅5⋅6×3⋅4+2⋅3⋅4⋅5×4⋅5+1⋅2⋅3⋅4×5⋅6=4!2⋅11!7!4! 274 Generally, if the given series be 𝑃₁𝑄₁+𝑃₂𝑄₂+⋯+𝑃_𝑛−1𝑄_𝑛−11 where 𝑄_𝑟=𝑟(𝑟+1)(𝑟+2)⋯(𝑟+𝑞−1) and 𝑃_𝑟=(𝑛−𝑟)(𝑛−𝑟+1)⋯(𝑛−𝑟+𝑝−1) the sum of 𝑛−1 terms will be 𝑝!𝑞!(𝑝+𝑞+1)!⋅(𝑛+𝑝+𝑞−1)!(𝑛−2)!275

Miscellaneous Series

Sum of thepowers of the terms of an Arithmetical Progression

1+2+3+⋯+𝑛=𝑛(𝑛+1)2=𝑆₁
            1+2²+3²+⋯+𝑛²=𝑛(𝑛+1)(2𝑛+1)6=𝑆₂
            1+2³+3³+⋯+𝑛³=𝑛(𝑛+1)(2𝑛+1)6²=𝑆₃
            1+2⁴+3⁴+⋯+𝑛⁴=𝑛(𝑛+1)(2𝑛+1)(3𝑛²+3𝑛−1)30=𝑆₄

By the method of Indeterminate Coefficients (234). A general formula for the sum of the 𝑟^th powers of 1⋅2⋅3⋯𝑛, obtained in the same way is 𝑆_𝑟=1𝑟+1𝑛^𝑟+1+12𝑛^𝑟+𝐴₁𝑛^𝑟−1+⋯+𝐴_𝑟−1𝑛 where 𝐴₁, 𝐴₂, ⋯ are determined by putting 𝑝=1, 2, 3, ⋯ successively in the equation 12(𝑝+1)! =1(𝑝+2)!+𝐴₁𝑟(𝑝)!+𝐴₂𝑟(𝑟−1)(𝑝−1)!+⋯+𝐴_𝑝𝑟(𝑟−1)⋯(𝑟−𝑝+1)! 276 𝑎^𝑚+(a+𝑑)^𝑚+(a+2𝑑)^𝑚+⋯+(a+𝑛𝑑)^𝑚=(𝑛+1)𝑎^𝑚+𝑆₁𝑚𝑎^𝑚−1𝑑+𝑆₂𝐶(𝑚,2)𝑎^𝑚−2𝑑²+𝑆₃𝐶(𝑚,3)𝑎^𝑚−3𝑑³+⋯ Proof. By Binomial Theorem and (276).277

Summation of a series partly Arithmetical and partly Geometrical

Example

To find the sum of the series 1+3𝑥+5𝑥²+⋯ to 𝑛 terms. Let


        𝑠=1+3𝑥+5𝑥²+7𝑥³+⋯+2𝑛−1)𝑥^𝑛−1
        𝑠𝑥=  𝑥+3𝑥²+5𝑥³+⋯+(2𝑛−3)𝑥^𝑛−1+(2𝑛−1)𝑥^𝑛

∴ by subtraction, 𝑠(1−𝑥)=1+2𝑥+2𝑥²+2𝑥³+⋯+2𝑥^𝑛−1−(2𝑛−1)𝑥^𝑛 =1+2𝑥1−𝑥^𝑛−11−𝑥−(2𝑛−1)𝑥^𝑛 ∴ 𝑠=1−(2𝑛−1)𝑥^𝑛1−𝑥+2𝑥(1−𝑥^𝑛−1)(1−𝑥)² 278 A general formula for the sum of 𝑛 terms of 𝑎+(a+𝑑)𝑟+(a+2𝑑)𝑟²+(a+3𝑑)𝑟³+⋯ is 𝑆=𝑎−(a+~~𝑛−1~~𝑑)𝑟^𝑛1−𝑟+𝑑𝑟(1−𝑟^𝑛−1)(1+𝑟)² Obtained as in (278) Rule. Multiply by the ratio and subtract the resulting series. 279 11−𝑥=1+𝑥+𝑥²+𝑥³+⋯+𝑥^𝑛−1+𝑥^𝑛1−𝑥 280 1(1−𝑥)²=1+2𝑥+3𝑥²+4𝑥³+⋯+𝑛𝑥^𝑛−1+(𝑛+1)𝑥^𝑛−𝑛𝑥^𝑛+1(1−𝑥)²281 (𝑛−1)𝑥+(𝑛−2)𝑥²+(𝑛−3)𝑥³+⋯+2𝑥^𝑛−2+𝑥^𝑛−1=(𝑛−1)𝑥−𝑛𝑥²+𝑥^𝑛+1(1−𝑥)² By (253) 282 1+𝑛+𝑛(𝑛−1)2!+𝑛(𝑛−1)(𝑛−2)3!+⋯=2^𝑛 1−𝑛+𝑛(𝑛−1)2!−𝑛(𝑛−1)(𝑛−2)3!+⋯=0 By making 𝑎=𝑏 (125) 283 The series 1-𝑛−32+(𝑛−4)(𝑛−5)3!−(𝑛−5)(𝑛−6)(𝑛−7)4!+⋯+(−1)^𝑟−1(𝑛−𝑟−1)(𝑛−𝑟−2)⋯(𝑛−2𝑟+1)𝑟! consists of 𝑛2 or 𝑛−12 terms, and the sum is given by 𝑆=3𝑛if 𝑛 be of the form 6𝑚+3, 𝑆=0if 𝑛 be of the form 6𝑚±1, 𝑆=−1𝑛if 𝑛 be of the form 6𝑚, 𝑆=2𝑛if 𝑛 be of the form 6𝑚±2, Proof: By (545), putting 𝑝=𝑥+𝑦, 𝑞=𝑥𝑦, and applying (546) 284 The series 𝑛^𝑟−𝑛(𝑛−1)^𝑟+𝑛(𝑛−1)2!(𝑛−2)^𝑟−𝑛(𝑛−1)(𝑛−2)3!(𝑛−3)^𝑟+⋯ takes the values 0, 𝑛!, 12𝑛(𝑛+1)! according as 𝑟 is <𝑛, =𝑛, or =𝑛+1 Proof: By expanding (𝑒^𝑥−1)^𝑛, in two ways: first, by the Exponential Theorem and Multinomial; secondly, by the Bin. Th., and each term of the expansion by the Exponential. Equate the coefficients of 𝑥^𝑟 in the two results. Other results are obtained by putting 𝑟=𝑛+2, 𝑛+3, ⋯. The series (285), when divided by 𝑟!, is, in fact, equal to the coefficient of 𝑥^𝑟 in the expansion of ~~𝑥+𝑥²2!+𝑥³3!+⋯~~^𝑛 285 By exactly the same process we may deduce from the function {𝑒^𝑥−𝑒^−𝑥}^𝑛 the result that the series 𝑛^𝑟−𝑛(𝑛−2)^𝑟+𝑛(𝑛−1)2!(𝑛−4)^𝑟−⋯ takes the values 0 or 2^𝑛⋅𝑛!, according as 𝑟 is <𝑛 or =𝑛; this series, divided by 𝑟!, being equal to the coefficient of 𝑥^𝑟 in the expansion of 2^𝑛~~𝑥+𝑥³3!+𝑥⁵5!+⋯~~^𝑛 286