excess of the first term above the last, or equal to the first term, barely; because the last is supposed to vanish, or to be indefinitely small in respect of the first. Hence it appears that 1-b+b×1—c+be × 1-d+bed × 1-e+bcde × 1—ƒ, &c. = 1.

X

m a -m m m+p a-m m + X + X ·+. X a a+p a a+p a+q a

m+pm+q -m

a+p a + q

m

X + &c. = 1; and consequently 1+

a+r

+

X

X

m m+p m m+ pm + q a+p a+p a+q a+pa+q a + r

a

a-m

a- m by dividing the whole by

a

+ &c. =

Hence, if q be taken =2p, r=3p, s=4p, &c. and 3 be

+

n

1

+ &c.

2 n

;

exhibiting the

1.2.3.4

n.n+1.n+2.n+3 general value of a series of the reciprocals of figurate numbers, infinitely continued; whereof the order is represented by n from whence as many particular values as you please may be determined. Thus, by expounding n by 3, 4, 5, &c. successively, it appears that

And so on, for any higher order; but the sums of the two first, or lowest orders, cannot be determined, these being infinite.

By interpreting and m by different values, the sums of various other series may be deduced from the same general equation. Thus, in the first place, let ß =

m+ 2; so shall the said equation become 1 +

m.m+1

+

+

&c.

m.m + 1 m.m + 1 m + 2.m + 3 m + 3.m + 4 m + 4.m + 5 =m+1; which, divided by m. m + 1, gives

Again, by taking @ = m + 3, and dividing the whole equation by m. m + 1. m +2, we have

From whence the law for continuing the sums of these last kinds of series is manifest; by which it appears, that, if instead of the last factor in the denominator of the first term, the excess thereof above the first factor be substituted, the fraction thence arising will truly express the value of the whole infinite series.

A few other particular cases will further show the use of the general equations above exhibited.

2

2
X

4

2

Let the sum of the series 1++ × 7+ 5 ×

5 5

Here, by comparing the proposed series with 1 +

1

+

and p= 2; and consequently

value of the series.

Let the sum of an infinite series of this form, viz.

1.2.

1

3 &c. 2.3.4 &c.

manded.

If, instead of the whole infinite series, you want the sum of a given number of the leading terms only; then let the value of the remaining part be found, as above, and subtracted from the whole, and you will have your desire. Thus, for instance, let it be required to find the sum of

1

&c. Then the remaining part, 11.12 +

1

+

11 (by the rule above)

and the whole series = 1, the value here sought will there

The sum of series arising from the multiplication of the terms of a rank of figurate numbers into those of a decreasing geometrical progression, are deduced in the following manner.

By the theorem for involving a binomial (given at p. 40. and demonstrated hereafter) it is known that

In which equation let m be expounded by 1, 2, 3, 4, 5, &c. successively, so shall

= 1 + x + x2 + 203 + x2 + x2 + &c.

= 1 + 2x + 3x2 + 4x3 + 5x4 + 6x3, &c.

= 1 + 3x+6x2+10x2+15xa+21x3, &c.

=1+4x+10x2+20x3+35x+56x5, &c.

= 1 +5x+15x2+35x3 +70x1+126x3, &c.

1+6x+21x2+56x+126x1+252x3, &c.

All which series (whereof the sums are thus given) are ranks of the different orders of figurate numbers, multiplied by the terms of the geometrical progression, 1, x, x2, x3, x4, &c.

From these equations the sums of series composed of the terms of a rank of powers, drawn into those of a geometrical progression, such as 1+4x+9x2+16x3, &c. and 1 +8x+27x2+64x3, &c. may also be derived; there being, as appears from the former part of this section, a certain relation between the terms of a series of powers and those of figurate numbers; the latter being there determined by means of the former. To find here the converse relation, or to determine the former from the latter, it will be expedient to multiply the several equations above brought out, by a certain number of terms of an assumed series 1+Ax+Bx2+Ca3, &c. in order that the coefficients of the powers of x may, by regulating the values A, B, C, D, &c. become the same as in the series given.

Thus, if the series given be 1+4x+9x2+16x3+25x, &c.; then, by multiplying our third equation by 1+Ax, 1+ Ax

we shall have3≈ 1 + 3 + A × x2+6+ SA×x2+

10+6Axx3+&c. which series, it is evident by inspection, will be exactly the same, in every term, with the proposed one, if the quantity A be taken = 1. 1. The sum of the said series, infinitely continued, is therefore truly repre

1 + 4x + 10x2 + 20x3 + 35x, &c. be multiplied by