and the number of terms will be finite if v be rational and integral.

The series in (6) comes from the equivalence of the operations denoted by the symbols E and (1A). In like manner we may obtain a cognate expression from the equivalence of E and (1+4). This gives us, when we perform them on † (x),

x (x + 1)

$(0) = (x) – − x. A(x) +∞ (α

1.2

▲3p (x) - &c.

Putting as before 4(x)=Σv, and C for $(0), and transpos

ing, we get

Σv2 = C+ xv2

x(x+1)
1 .2

Av2+ &c.

.....

(8)*.

In applying the above to the summation of series we may avoid the use of an undetermined constant and render the demonstration more direct by proceeding as follows: Va+Ya+s+...+Va+x-1= {1 + E + E2 + ... + Ex1 } va

* That the constants in (7) and (8) are the same appears evident when we consider that (8) may be obtained from (7) by mere algebraical transformation. The series-portions are in fact the results of performing the equivalent (1+A)* −1 1 − (1+A)-* and Ex on vo

direct operations

A

Here all the operations performed on v. are direct, and the result is given in differences of the first term.

Ex. 3. To find the sum of x terms of the series 1'+2+... Applying* (7) we have (since ▲v=1, A3⁄4v = 2)

-

1′+ 2′+ ... + (x − 1)'= Σœ2= C+≈ (x − 1) + x (x − 1) (x − 2) 2.

=

2

1.2.3

Putting 2 we see that C is zero, and adding a2 to both sides we obtain

Ex. 4. Find the sum of n terms of the series whose nth term

In practice it will be found better to resolve the nth term into factorials

and apply the rule given in the note to page 68.

u, being of the form ax + b, and ☀ (x) a rational and integral function of x of a degree lower by at least two unities than the degree of the denominator.

Expressing (x) in the form

$(x)=A+Bu2+ Cu2ux+1+...+EU2Ux+1 ••• Ux+m-2

...

A, B... being constants to be determined by equating coefficients, or by an obvious extension of the theorem of Chap. II. Art. 5, we find

and each term can now be integrated by (5).

Again, supposing the numerator of a rational fraction to be of a degree less by at least two unities than the denominator, but intermediate factors alone to be wanting in the latter to give to it the factorial character above described, then, these factors being supplied to both numerator and denominator, the fraction may be integrated as in the last case.

Ex. 5. Thus u still representing ax + b, we should have

with the second member of which we must proceed as before.

Ex. 6. Find the sum of n terms of the series

+

n (n + 1) (n + 2) (n + 3) ̄ ̄ (n + 2) (n + 3)

1

+

1

(n + 1) (n + 2) (n+3) n (n + 1) (n + 2) (n + 3) '

The sum of n terms therefore, by the rule on page 68,

We thus can find the sum of n terms of any series whose nth term is (n), provided that (n) be either (1) a rational integral function of n, or (2) a fraction whose denominator is the product of terms of an arithmetical series that remain a constant distance from the nth term, and whose numerator is of a degree lower by at least two than its denominator*

5th Form. Functions of the form a* or a p(x) where (x) is rational and integral.

* Since 4(n)en* =4(D) ɛnx we may write

(a)+(a+1)+... (a + n − 1) = [$ (D) {e2 + ε(a+1)~ + e(ata-1}]x-0

and the series may therefore be summed by the methods of Differential Calculus or Differential Equations according as (n) is an integral function of n or not. That the result thus obtained is identical with that in the text follows from the identity demonstrated in (16) page 23, viz.

={(a+n)-(a)}=Σp(a+n) -Σp(a),

which agrees with the previous expression.

a*

From (13) page 8, we obtain at once Σa=1. For the integration of a p (x) we shall have recourse to symbolical methods.

= a* (α€” − 1) ̄1μ (x) = a* {a (1 + ▲) — 1} ̄1¤ (x)

1

(1 +

1)(x)

to which of course an arbitrary constant must be added.

It will be found that the direct application of this theorem + is the simplest method of summing such series as have their ath term of the form a*. 4 (x).

* By means of the well-known formula f(D)emx p(x) = εmx ƒ (D+m) $ (x). The proof of this formula is given in Boole's Diff. Eq. (First Ed., p. 385), and in many other books.

+ The demonstration of (10) can be still farther simplified by quoting the theorem,

ƒ (E) a* p (x)=a*ƒ (aE) $ (x).

This may be deduced from the formula above quoted, but is more readily demonstrated independently, since if An En be one term of the expansion of ƒ (E) in powers of E we have

An E ap(x)=A„a*+ p (x+n)=a* . Aña1 Eˆ p(x)=a*. A„(aE)" p(x), summing all such terms we get

f(E) ap(x)=a*ƒ (aE) $ (x),

and the demonstration of (10) runs thus,

A-1 ap (x)=(E-1)-1 a*p(x) = a*(aE−1)−1 p(x)

=a* {a (1+4)-1}(x)=&c.