Elements of Algebra

168 ELEMENTS SECT. III. 517. When the exponent is less than 1, and, consequently, when the terms of the progression continually diminish, the sum of such a decreasing progression, carried on to infinity, may be accurately expressed. For example, let the first term be 1, the ratio $\frac{1}{2}$, and the sum $s$, so that: $s = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \ldots$ ad infinitum. If we multiply by 2, we have $2s = 2 + 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots$ ad infinitum: and, subtracting the preceding progression, there remains $s=2$ for the sum of the proposed infinite progression. 518. If the first term be 1, the ratio $\frac{1}{3}$, and the sum $s$; so that $s = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81} + \ldots$ ad infinitum: Then multiplying the whole by 3, we have $3s = 3 + 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots$ ad infinitum; and subtracting the value of $s$, there remains $2s=3$; wherefore the sum $s = 1\frac{1}{2}$. 519. Let there be a progression whose sum is $s$, the first term 2, and the ratio $\frac{3}{4}$; so that $s = 2 + \frac{3}{2} + \frac{9}{8} + \frac{27}{32} + \frac{81}{128} + \ldots$ ad infinitum. Multiplying by $\frac{4}{3}$, we have $\frac{4}{3}s = \frac{8}{3} + 2 + \frac{3}{2} + \frac{9}{8} + \frac{27}{32} + \frac{81}{128} + \ldots$ ad infinitum; and subtracting from this progression $s$, there remains $\frac{1}{3}s = \frac{8}{3}$: wherefore the sum required is 8. 520. If we suppose, in general, the first term to be $a$, and the ratio of the progression to be $\frac{b}{c}$, so that this fraction may be less than 1, and consequently $c$ greater than $b$; the sum of the progression, carried on ad infinitum, will be found thus: Make $s = a + \frac{ab}{c} + \frac{ab^2}{c^2} + \frac{ab^3}{c^3} + \frac{ab^4}{c^4} + \ldots$ Then multiplying by $\frac{b}{c}$, we shall have $\frac{b}{c}s = \frac{ab}{c} + \frac{ab^2}{c^2} + \frac{ab^3}{c^3} + \frac{ab^4}{c^4} + \ldots$ ad infinitum; and subtracting this equation from the preceding, there remains $(1 - \frac{b}{c})s = a$.