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then unity is the limit of the sum, since the sum of any number of its terms is less than unity; and since no number can be added to the sum of a certain number of its terms, but what must make the sum more than a unit.

3. The sum of an infinite series is the limit of the sum of that series. 4. A series is said to be summable when a finite general expression can be found, that will be the value of a given number of its terms, or when a limit to the sum can be found when the number of the terms is infinite; otherwise the series is not summable.

Theorem.

C.

The integral of the (x+1)th term of a series, will be equal to the sum of all the preceding r terms of that series. For let there be the series a, b, d, &c. Then the successive sums are a, a+b, a+b+c, a+b+c+d, &c. Now since any term of the upper series is the difference of the corresponding term of the lower series, and the preceding term of the said lower series; therefore, calling the upper series the series of differences, and the lower series the series of sums, any term, as x+1, in the series of differences, will be the difference between the rth term and the (x+1)th term in the series of sums; but what is called the difference of a function of x, is another function of x+1, which is the difference between the first function of x, and the same function as the first of x+1: therefore, reciprocally, since a certain function of x, is the integral of a difference which is a function of x+1, the integral of the (x+1)th term in the series of differences will be the rth term in the series of sums; that is, the sum of x terms of the series of differences. Q.E.D.

Problem.

To find the sum of a given number of terms of a series, the law of the series being given,

Let x represent the number which indicates the number of the terms. Find the general term of the series, and the integral of the (x+1)th term by such of the former rules as applies, then the integral corrected, if need be, is the sum of the series.

N. B. If the general term of the series is not expressed factorials, it will be most convenient to reduce it to factorials, and this may be done by Problem 1, when the genera! term is formed of binomial factors.

Notation.

Let represent the rth term, y the (x+1)th term of the series;

also let ▲ denote the integral of y, and E the integral corrected, or the sum required

Examples.

1. Find the sum of 100 ternis of the series, 1+2+3+&c

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2. Find the sum of 50 terms of the series 1.2.3+2.3.4+3.4.5+

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3. Find the sum of 20 terms of the series 12+22+3*+4o+&c. Here t= x2=21-2 by decomposition,

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4. Find the sum of x terms of a series, whose general term is represented by ar3+bx2+cx+d.

By decomposition ax3 (x3-3x2 +2111),

bx2=b(...... x2-xx1),

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..tax31+(b-3a) x2+(a−b+c)x111+d,

y=a(x+1)+(b−3a)(x+1)212 +(a−b+c)(x+1)+d(x+1)TM,

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X311

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2.1

Ay=a- +、 −3a) 3,11 +(a−b+c) 2011 +dz11, correct;

whence,

4.1

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2311

x211

Στα + (b−3 a) 12 + (a − b + c ) 221 +dx.

4

3

More Examples for Fractice.

5. Find the sum of 15 terms of the series 53+79+93+&c. 6. Find the sum of a terms of the series of products 1.2'+3'4'+5°.6+79.8+&c.

L

7. Find the sum of x terms of the series of products

12.23 +223 +39.43 +42.53+ &c. Ans.

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2x311

3

8. Find the sum of x terms of the series of products

x41x311 211
X

*.32+2.52+3.72+4.92+5.119+&c. Ans. + +

4 3

2

9. Find the sum of x terms of the series of products 1.2.3+2.4.6+3.69+4.8.12+ &c. as also the sum of ten terms of

the same series. Ans.

3x41

2

-x311+8x211.

10. Find the sum of 16 terms of the series of products

3.4.6+5.5.9+7.6.12+ &c. Ans. 3

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3x211

+231 + +3x

2

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=

12. Find the sum of ten terms of the series of fractions

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ду

(x+2)-211

-2

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-1

2(x+2) 211

-1

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series; we may, therefore, find the sum of an infinite series of fractions, by the following

Rule.

Transfer the denominator to the numerator, and the exponent will become negative; add unity to the exponent, and divide the fraction by the exponent thus increased, and by the difference, and the fraction thus divided, will be the sum of the infinite series.

Observing in dividing by the negative exponent-to change its sign into +.

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Scholium. When the numerators contain any powers or products of the distance, with or without constant quantities, the integral will be easily found by the following

Rule.

Take the differences of the numerators of the given fractions until the last order became constant: then the first numerator of the given fractions, the first of the first, the first of the second, the first of the third, &c. order of differences are the respective numerators of the integrals; find as many orders of the integrals of the respective derominators as the numerators so found, and these integrals will be the respective denominators of the fractions, which will be the value of the sum.

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Here the numerators are a+b, u+2b, a+3b, &c. and their first differences are

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ს,

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ს,

148

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Second differences,

5,

7,

2, constant.

Then as many of the succeeding values of the denominators ·

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1 1

1

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2

Whence Σ=

T=1+ (r−1) =+ (r−1).

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