an infinite series by actually dividing the numerator 1 by the denom 258. If we make a 1, we have this remarkable comparison: 1 =} = 1−1+1−1+1− 1 + 1 − 1, &c. to ini+a finity. This will appear rather contradictory; for if we stop at -1, the series gives 0; and if we finish by + 1, it gives 1. But this is precisely what solves the difficulty; for since we must go on to infinity without stopping either at 1 or at +1, it is evident that the sum can neither be 0 nor 1, but that this result must lie between these two, and therefore be }. 259. Let us now make a = 1, and our fraction will be which must therefore express the value of the series +, &c. to infinity. If we take only the two leading terms of this series, we have, which is too small by . If we take three terms, we have, which If we take four terms, we have §, which is too is too much by small by, &c. 260. Suppose again a; our fraction will be = 1 and to this the series 1-+-+-+15, &c. continued to infinity, must be equal. Now, by considering only two terms, we have, which is too small by. Three terms make 1, which is too much by Four terms make, which is too small by, and so on. 1 261. The fraction may also be resolved into an infinite 1+a series another way; namely, by dividing 1 by a + 1, as follows: And if we suppose a = 2, we shall have the series 1 −1 +1-76 + 32 - 6's, &c. — ¦. 262. In the same manner, by resolving the general fraction into an infinite series, we shall have, C a+b bc a S༐s S|8 C Whence it appears, that we may compare with the series a+b &c. to infinity. Let a = 2, b = 4, c= 3, and we shall have 3 = a+b 2+4 ====3+6-12, &c. Let a 10, b = 1, and c = 11, and we have If we consider only one term of this series, we have 11, which is too much by; if we take two terms, we have, which is too small by; if we take three terms, we have 188, which is too much by To, &c. 1 263. When there are more than two terms in the divisor, we may also continue the division to infinity in the same manner. series to which it is equal would be found as follows: 1 1-a+aa =1+a—a3 — a1 + a® + a2 — ao — a1o, &c. Here, if we make a 1, we have 1=1+1—1—1+1+1—1—1+1+1, &c. which series contains twice the series found above, 1-1+1-1+ 1, &c. Now, as we have found this = 1, it is not astonishing that we should find, or 1, for the value of that which we have just determined. Make a , and we shall then have the equation 1 ++ ==1+- 1 — 18 + os + 15 - 87, &c. Suppose a = 1, we shall have the equation If we take the four leading terms of this series, we have, which is only, less than . Suppose again a = &, we shall have This series must therefore be equal to the preceding one; and sub 264. The method which we have explained, serves to resolve, generally, all fractions into infinite series; and, therefore, it is often found to be of the greatest utility. Further, it is remarkable, that an infinite series, though it never ceases, may have a determinate value. It may be added, that from this branch of methematics inventions of the utmost importance have been derived, on which account the subject deserves to be studied with the greatest attention. CHAPTER VI. Of the Squares of Compound Quantities. 265. WHEN it is required to find the square of a compound quantity, we have only to multiply it by itself, and the product will be the square required. For example, the square of a + b is found in the following man ner: a+b aa + ab ab+bb aa +2ab+bb. 266. So that, when the root consists of two terms added together, as a + b, the square comprehends, 1st, the square of each term, namely, a a and bb; 2dly, twice the product of the two terms, namely, 2 ab. So that the sum a a+2ab+bb is the square of a+b. Let, for example, a = 10 and 6 = 3, that is to say, let it be required to find the square of 13, we shall have 100 + 60 + 9, or 169. |