Elements of Algebra

CHAP. V. OF ALGEBRA. 91 292. This being the case, we may continue the series as far as we please, without being under the necessity of performing any more calculations; and thus we shall have $\frac{1}{1-a} = 1+a+a^2+a^3+a^4+a^5+a^6+a^7+\frac{a^8}{1-a}$; or we might continue this farther, and still go on without end; for which reason it may be said that the proposed fraction has been resolved into an infinite series, which is, $1+a+a^2+a^3+a^4+a^5+a^6+a^7+a^8+a^9+a^{10}+a^{11}+a^{12}, \ldots$ to infinity: and there are sufficient grounds to maintain, that the value of this infinite series is the same as that of the fraction $\frac{1}{1-a}$. 293. What we have said may at first appear strange; but the consideration of some particular cases will make it easily understood. Let us suppose, in the first place, $a=1$; our series will become $1+1+1+1+1+1+1, \ldots$; and the fraction $\frac{1}{1-a}$, to which it must be equal, becomes $\frac{1}{1-1}$, or $\frac{1}{0}$. Now, we have before remarked, that $\frac{1}{0}$ is a number infinitely great; which is therefore here confirmed in a satisfactory manner. See Art. 83 and 84. Again, if we suppose $a=2$, our series becomes $1+2+4+8+16+32+64, \ldots$ to infinity, and its value must be the same as $\frac{1}{1-2}$, that is to say $\frac{1}{-1}=-1$; which at first sight will appear absurd. But it must be remarked, that if we wish to stop at any term of the above series, we cannot do so without annexing to it the fraction which remains. Suppose, for example, we were to stop at $64$, after having written $1+2+4+8+16+32+64$, we must add the fraction $\frac{128}{1-2}$ or $\frac{128}{-1}$, or $-128$; we shall therefore have $127-128$, that is in fact $-1$. Were we to continue the series without intermission, the fraction would be no longer considered; but, in that case, the series would still go on. 294. These are the considerations which are necessary, when we assume for $a$ numbers greater than unity; but if we suppose $a$ less than $1$, the whole becomes more intelligible: for example, let $a=\frac{1}{2}$; and we shall then have $\frac{1}{1-a} = \frac{1}{1-\frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2$, which will be equal to the following series $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}, \ldots$ to infinity.