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2.7311

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

2011

arsit 10x411 19.29 + 22 39 + 38.43 +42.59 + &c. Ans.

+
6 5 4

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

2411 311 ..3 +2.52 +3.72 +4.9° +5.11% +&c. Ans. + +

4 3 2 9. Find the sum of x terms of the series of products 1.2.3+2.4.6+3.6.9+4.8.12+&c. as also the sum of ten terms of

3.3411 the same series. Ans. -2311 + 8x21.

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

3411 3.4.6+5.5.9+7.6.12+ &c. Ans. 3 + 2311 +

-+3.7

2 11. Find the sum of ten terms of the series

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3'+3+33 +3 +35+ &c. Ans.

2. 12. Find the sum of ten terms of the series of fractions 1 1

1 + +

+ &c. 2.3.4 3.4.5.4.5.6

1 Here t =

1
y=(x+2)-31
(x+2)-2!1

-1
Ду
-2 2(x+2) 211
-

1 Make x=0, then will

whence it appears 2(x + 2)211

2.2211

2.2211

126

(

2.1297

2 (2211

1241)

that the integral is too small by which, therefore, must be

2.2211

1 added to it; and we shall have a

Now 2(x+ 2)211 •

1 1 1 1 make x = 10, and we shall have

2.2211 for the sum of ten terms.

1 .

is the sum of the infinite

2.2017 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 siga into t.

Examples.
1

1 .. Find the sum of the Infinite series +++

+&c. 1.3' 3.5 5.7

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+

the sum.

3. Find the sum of the infinite series

1

1

+ &c.
3.6.9.12 6.9.12.15
1
Here

=3-413.
3.6.9.12
3-313

1
A3-415 =

9 9.3.6.9 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 deriominators as the numerators so found, and these integrals will be the respective denominators of the fractions, which will be the value of the sum.

Examples i Find the sum of the infinite series a+26

a +36

+ &c.
(m + r) 3lr+m+2r),
Here the numerators are a+1, +2b, a +36, &c.
and their first differences are

b, b,
Now A31r
S'm31

2r

2+6

m -21

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Numerators 4,

9, 16, First differences, 5, 7,

Second differences, 2, constant. Then as many of the succeeding values of the denominators T 1 1

1

1
and their integrals are

(-1)
4
5

2 Whence =

+

+
(1-1)** (r-in

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ON THB

CONVERGENCY OF SERIES.

Problem 1. To find the convergency of an Infinite Series in general terms, x being the distance of the term from the beginning.

Find the factors in terms of 7, the factors or variable quantities being supposed to be in arithmetical progression; then multiply into, or divide by the constant quantities, will give the general term, which reserve; then form the preceding term in the same manner, by writing r-1 for I; divide the subsequent term by the preceding; and the quotient will shew the convergency of the series for any value of x.

Examples. 1. To find the convergency of the infinite series 1

1
+
noa

mots
rad

na

1

+

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n!

nina 1 therefore

which is the convergency roquired, and is in

nua this example a constant quantity. 2. To find the convergency of the infinite series

1 1 1 1 1

+ +

1

2

3

4

.... The preceding term is –,; therefore the convergency is * !

3. To find the convergency of the infinite series

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(2-2)ml? (2-2)-112 2–18 [z+?(m-1)]

z+z(m-1)
Examples 2 and 3 are both included in this.
5. Find the convergency of the series

1
1

1
1-
9.39 13.36.

1 Here the general term is

r»[4(2-1)+1]; or, putting a for the first factor, and c for the increment, it will be

and the gru 1)'

5.38't

5–1th term will be *~*(012–2)+a); whence, dividing the s

for the convergency

c(-2+a term by the I-1th term, gives

ra[c(2-1)+a] of the series.

Scholium. Letx=2; then the convergency

c(1-2)+a pre [c(x - 1) + a)

becomes simply mic4a) · Whence, if the first factor a be increased,

a

(ca the convergency is less ; that is to say, supposing med and c to be con stant, the greater a is, the nearer will the expression

pe (c+a) approach to which is its limit ; and if r=1, this limit wouid be unity. 6. Find the convergency of the infinite series

m21102

+ &c. R?pm+116 Ropm+2! Ropm+ 3le

.) This series is equivalent to тся

(m+1)CB Romic

R(P+mc) * (+(m+1)x)+ &c. Therefore the con

mc

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+

1

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