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ON THE

CONVERGENCY OF SERIES.

Problem 1.

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

Find the factors in terms of x, 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 -1 for ; divide the subsequent term by the preceding; and the quotient will shew the convergency of the series for any value of x.

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this example a constant quantity.

2. To find the convergency of the infinite series

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=

' [z+3(m−1)] ~ z+z(m−1)

Examples 2 and 3 are both included in this.

5. Find the convergency of the series

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1

ro[c(x−1)+a]'

x-1th term will be ~~(~(x-2)+a]; whence, dividing the

and the

term by the z-1th term, gives

c(x-2+a
ra[c(x-1)+a]

for the convergency

of the series.

Scholium. Let x=2; then the convergency

a

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c(x-2)+a rd[c(x − 1) + a]

becomes simply(c+a) Whence, if the first factor a be increased,

the convergency is less; that is to say, supposing and c to be con

a

stant, the greater a is, the nearer will the expression (c+a)

a

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approach to = which is its limit; and if r=1, this limit

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vergency at the rth term is expressed by

c[m+x-2]
R[p+(m+x-2)c]

Scholium. If x=2, then the convergency

c(m+x-2)
R[p+(m+x-2)c]

be

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in the second term; whence it appears, that, by

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 ;, 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 r; 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

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2

are constant, therefore the limit of convergency is = 1; which

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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 be less than the -1th 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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3. Find the limits of convergency of the infinite series

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9(a+4); which is the

convergency at the second term,

4. Find the limit of convergency of the infinite series

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

=1, which is the limit of convergency.

5. Find the limit of convergency of the infinite series

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Therefore the general convergency is; but, when x is infinitely

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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

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The series here proposed gives the number v, whose Napierian logarithm Vis 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

A'v (A'v)2, (A'v)3 (A'v)*

+

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+ &c.

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This series gives the logarithm of the number v+▲'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.

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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)s*3(1+v)3 l+v) TM 2 (1 +v) s *3 ( 1 +v)s+&c. to find

the limit of convergency.

Here the general convergency is

convergency is

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= ; so that this series will always con(1+v)x 1+v

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

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