We have thus divided all the series, except the first two terms, into lots, each containing half as many terms as there are units in the denominator of its last term. Thus, the fourth lot contains 16 or ³²/₂2 terms. Each of these lots may be shewn to be greater than ½. Take the third, for example, consisting of ⅑, ¹/₁₀, ¹/₁₁, ¹/₁₂, ¹/₁₃, ¹/₁₄, ¹/₁₅, and ¹/₁₆. All except ¹/₁₆, the last, are greater than ¹/₁₆; consequently, by substituting ¹/₁₆ for each of them, the amount of the whole lot would be lessened; and as it would then become ⁸/₁₆, or ½, the lot itself is greater than ½. Now, if to 1 + ½, ½ be continually added, the result will in time exceed any given number. Still more will this be the case if, instead of ½, the several lots written above be added one after the other. But it is thus that the series 1 + ½ + ⅓, &c. is composed, which proves what was said, that this series has no limit. 198. The series 1 + _r_ + _rr_ + _rrr_ + &c. always has a limit when _r_ is less than 1. To prove this, let the term succeeding that at which we stop be _a_, whence (194) the sum is 1 - _a_ 1 _a_ -------, or (112) ------- - ------. 1 - _r_ 1 - _r_ 1 - _r_ The terms decrease without limit (196), whence we may take a term so far distant from the beginning, that _a_, and therefore _a_ -------, 1 - _r_ shall be as small as we please. But it is evident that in this case 1 _a_ ------- - ------- though always less than 1 - _r_ 1 - _r_ 1 1 -------- may be brought as near to ------- 1 - _r_ 1 - _r_ as we please; that is, the series 1 + _r_ + _rr_ + &c. continually approaches to the limit 1 --------. 1 - _r_ Thus 1 + ½ + ¼ + ⅛ + &c. where _r_ = ½, continually approaches to 1 ----- or 2, as was shewn in the last article. 1 - ½ EXERCISES. 2 2 The limit of 2 + --- + --- + &c. 3 9 1 1 or 2(1 + --- + --- + &c.) is 3 3 9 9 81 ... 1 + --- + ---- + &c. ... 10 10 100 15 45 ... 5 + ---- + ---- + &c. ... 8¾ 7 49 199. When the fraction _a_/_b_ is not equal to _c_/_d_, but greater, _a_ is said to have to _b_ a greater ratio than _c_ has to _d_; and when _a_/_b_ is less than _c_/_d_, _a_ is said to have to _b_ a less ratio than _c_ has to _d_. We propose the following questions as exercises, since they follow very simply from this definition. I. If _a_ be greater than _b_, and _c_ less than or equal to _d_, _a_ will have a greater ratio to _b_ than _c_ has to _d_. II. If _a_ be less than _b_, and _c_ greater than or equal to _d_, _a_ has a less ratio to _b_ than _c_ has to _d_.