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cm

comes

cm +5-2] vergency at the oth term is expressed by

R[p+(m+x-2)c] Scholium. If x=2, then the convergency R[p+(m+1–2)c]

c(m+x-2)

bein the second term; whence it appears, that, by R(p+cm) increasing p, the first factor, the convergency will be greater ; but by increasing m, the number of factors, or C, the increment, the convergency will be less.

Problem 1 A series being given of which x is the number of the term from the beginning, to find the limit of convergency.

Find the convergency as in the last problem, and make the constant quantity or quantities in each of the factors that are not factors themselves, where added to or subtracted from x, equal to 0 in then multiply the factors as indicated together both in the numerator and denominator, and the resulting fraction will shew the limit of convergency.

This is evident, for a finite magnitude will not bear any proportion to an infinite one; therefore, considering x as infinite, the constant quantities may be looked upon as nothing when compared to the magnitude of I; and consequently the limiting convergency must be indicated by the fraction formed by the product of the remaining variable factors in the numerator and denominator into the constant multipliers.

Examples
1. Find the limit of convergency of the infinite series

1
+

to...t
(a + b) mulb (a+20)

1

+

Now the convergency is

%+ (m-1)

;

but since- :-x and z (m-1)

are constant, therefore the limit of convergency is

=l; which

indeed shews, that when x is infinite, the series has no convergency, unity being the standard of what may be so called ; for a series cannot be said to converge, unless the term at the distance x be less than the 11th term.

1 and 0 may be said to be the limits of convergency; the convergency being greater as the limit approaches to nothing, and less as it approaches to unity.

2. Find the limits of convergency of the infinite series

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The convergency of this series is

cm +1-2)
R[P+cm+1-2)]

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m

If x=2, then the first convergency will be Rip+zm)

3. Find the limits of convergency of the infinite series 1

1 1+

9.30+ + &c.

1

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13.30

gency is

.

a

4.3 +2-8 The convergency is

;

therefore the limit of conve9(4.0-4+a) 4.7 36x

4r+a-8 Make I=2, and

becomes

9(41-4+a)
convergency at the second term.
4. Find the limit of convergency of the infivite series
1 1

1 1
+ + + +
2"

4*

(x+1) The convergency is ; but when I is infinite, it will become

9(244); which is the

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3

+ &c.

=l, which is the limit of convergency.
5. Find the limit of convergency of the infinite series

1
1

1
it
+

+ 1".2" 1".2"3" 1*.2*.3".4" Or 1+an+B+C + D + BA Therefore the general convergency is ; but, when « is infinitely great, then the limit of convergency is 0. 6. Find the limit of the infinite series

1 1 1
it + + + +
am

a2m asm

att amm+&c.

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vergency is a constant quantity, and therefore the limit of the series is the same quantity.

7. Find the limit of convergency of the infinite series

V

Vs 1+ +

+ + &c. 1 1.2 1.2.3 1.2.3.4

+

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The series here proposed gives the number v, whose Napierian logarithm V is given, and consequently it will converge so that the limit will be nothing, or the convergency the greatest possible ; but a convergency will not take place until x - 1 be greater than V. 8. Find the limit of convergency of the infinite series An (A'v) (Av)' (A'v)*

303

Ad(1-1) The general convergency is ; therefore the limit of con

+

202

4ut +&c.

VX

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This series gives the logarithm of the number v+ A'v, by adding the sum of the series to the logarithm of v; and consequently will converge the more, the less Av is in proportion to v.

v va 9. Given the series om +

+&c. to find the limit of

2

3

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convergency.
The general convergency is

V(1—1)

;

whence the limit is And consequently this series diverges, unless v be a fraction less than unity.

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+ 10. Given the series 1(1+v)' 2(1 +v)' 3(1 +vs+&c. to find the limit of convergency.

V(x-1) Here the general convergency is ; therefore the limit of

(1+v).r

convergency is (1+uce itu

; so that this series will always con

verge, whatever be the value of v: but when v is great, the convergency will be very small, since v would be nearly=1+v.

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vi

vs 11. Given the infinite series 2 M

+
(0+2). '3(0+2)'

+ &c.) to find the limit of convergency. Here the general convergency is

v2 (25—3)

: therefore the lio

(v+2)'(2.0--1) 2 rv2 mit is

This series will therefore always con2x(v+2)*(0+2)2 verge, whatever be the value of v; but the greater v is, the less wili be the convergency.

This and the two preceding series have each the logarithm of vti for their value. The first converges the least, the ultimate rate being V; the second converges something more than the first, the ultimate

rate being j1 p) the last converges the most, the ultimate rate being

v2

In the first, v must be a fraction ; in the second and last (v+2)2 examples the series will always converge, whatever be the value of v.

v

vs 12. Given the infinite series 2

+

+ (2a+v) + v) s 52a + v

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(21-3) v2 Here the general convergency is

therefore the

(21-1)(2a +v): 2.xv? limit is

2x(2a + v) (2a + v) The value of this series added to the logarithm of a, gives the logarithm of atv.

The convergency may be increased in any degree, according as a is greater than v.

N.B. As the series in the last example may be derived from this by making a=1, so may also the limit of convergency be derived from the limit here found, and thus the rate of convergency will be lessened. 13. Find the limit of convergency of the infinite series 1.%v

1.23°o° 1.2.323v3
+

+ &c.
(1+0)
-v21)23+ )'

*

XV (2-1)
Here the general convergency is (I+v)[z+z(r1)]

whence

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V

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is the limit. When v is very great, approaches to unity, 1+v

Ito but never will equal it. If v be a fraction, the convergency will be greater, the greater the denominator is.

This series is a transformation of p-qutrve-V3+ &c. and is particularly useful in finding a rapid approximation to such slow con1 1 1

1 1 verging series as 1 - + at

+&c, or ) + +&c., where, in

3 5 7 both cases, v would be equal to unity, and 2 the first term of the series, z=1 in the first, and 2 in the second.

1

2

3

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being given, to find the limits of convergency.

v(n(x-2)-m) Here the general convergency is

na(x-1)

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Hence the convergency will be rapid when v is small

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compared with a : when r=3, the convergency in the third term is v( nm) 2na

The scries here proposed is the expansion of (a+u). Therefore to find the arithmetical expansion of (a+v)?, a ought to be a large square number, and v small. Thus, to find the square root of 5; then m=1,

m

n=2, make a=4, v=l, then añ=4' =2; the above series then be comes,

2(1+1.2.4

1
1
3

5 2 A

-C 1.2.4 2.2.4 3.2.4 4.2.4

+B.

+&c.)

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