in the progression, the smaller will be the quantity and am consequently the nearer the value of S will approach m from which it differs only by the quantity a a am (m 1) But it can never, strictly speaking, be equal to it, for the quantity will always have some value, however (m — 1) memi large n may be ; yet no quantity can be assumed, but this expression may be rendered smaller than it. The quantity is therefore the limit which the sum. 1 of a decreasing progression can never surpass, but to which the value continually approximates, as we take more terms in the series. In the progression 1, 3, 4, i, o, &c. 1 m m In this example the more terms we take, the nearer the sum of the series will approach to 2, but it can never be strictly equal to it. Now if we consider the number of terms infinite, 1 the quantity will be so small that it may be onitted 1 X 2nwithout any sensible error, and the sum of the series may be said to be equal to 2. By taking more and more terms we approach 2 thus, A 1 = 2-1 1+1 = 2-3 = 2-3 Examples. What is the sum of the series 1, $, , z't, &c. continued to an infinite number of terms ? S = 1 X 3 3 = 11. 3-1 2 2. What is the sum of the series, 5, 5, 6, 1t, &c. continued to an infinite number of terms ? 3. What is the sum of the following series continued to infinity ? 35, 7, 1, ', &c. 4. What is the sum of the following series continued to in finity ? 208, 26, 34, 1, &c. 5. What is the sum of the following series continued to infinity ? 38, 49, Hi,&c. 6. What is the 10th term of the series 5, 15, 45, &c.? 7. What is the sum of 8 terms of the series 35, 175, 875, &c.! When three numbers are in geometrical progression, the middle term is called a mean proportional between the other two. Let three numbers, a, b, c, be in geometrical progression, so that and b= (ac)* 8 Find a mean proportional between 4 and 9. 9. Find a mean proportional between 7 and 10. 10 Find a mean proportional betweerf 2 and 3 XLVIII. Logarithms. We have seen, Art. XXXVI, with what facility multiplication, division, the raising of powers, and the extraction of roots may be performed on literal quantities consisting of the same letier, by operating on the exponents. We propose now to apply the same principle, though in a way a little different, to nuinbers. Multiplication, we observed, is performed by adding the exponents, and division by subtracting the exponent of the divisor from that of the dividend. Thus a' X a® is au = d?. 'And is ao = is a?- = a'. In the same manner 20 x 24 = 2548 = 2, 27 and = 27-5 = 2. 2 Let us make a table consisting of two columns, the first containing the different powers of 2, and the second the exponents of those powers. Observe first that a = 1, so also 2° = 1, 2 = 2, 2* = 4, 2 = 8, 2* = 16, 28 = 32, 26 = 64, 2 = 128, &c. TABLE. Powers. | Expon. Powers. Expon. Powers. Expon. 1 0 128 7 16,384 14 2 1 256 8 32,768 15 4 2 512 9 65,536 16 8 3 1024 10 131,072 17 16 4 2048 11 262,144 18 32 5 4096 12 524,288 19 64 6 8192 13 1,048,576 20 Suppose now it is required to multiply 256 by 64. We find by the table that 256 is the 8th power of 2, that is 2, and that 64 is 2®. Now 28 x 24 = 2846 = 2". Returning to the table again and looking for 14 in the column of exponents, against it we find 16384 for the 14th power of 2. Therefore the product of 256 by 64 is 16384. This we may easily prove. 256 1024 1536 16384 Multiply 256 by 128. Finding these numbers in the table in the column of powers, and looking in the other column for the exponents, we find that 256 is the Sth power of 2, and 128 the 7th power. Adding the exponents 8 and 7, we have 15 for the exponent of the product. Now looking for 15 in the column of exponents, we find against it in the column of powers, 32768 for the 15th power of 2, which is the product of 256 by 128. Let the learner prove this by multiplying 256 by 128. Divide 8192 by 32. the corresponding exponents, we find 8192 is the 13th power of 2, and 32 is the 5th power, 9 = Looking for 8 in the column of exponents, and for its corresponding number, we find 256 for the 8th power of 2, or the quotient of 8192 by 32. Divide 32768 by 512. The exponents corresponding to these numbers in the table are 15 and 9. 15 :6. In the column of exponents, 6 corresponds to 64, which is the true quotient of 32768 by 512. What is the 3d power of 32 ? The exponent corresponding to 32 is 5. Now to find the 3d power of a we should multiply the exponent by 3, thus a5X3 = a'. So the third power of 2 is 25X3. = 26. Against 15 in the column of exponents we find 32768 for the 15th power of 2. Therefore the 3d power of 32 is 32768. What is the 2d power of 128 ? The exponent corresponding to this number is 7. 7 X 2 = 14. The number corresponding to the exponent 14 is 16384, which is the second power of 128. What is the 3d root of 4096 ? ej The number corresponding to the exponent 4 is 16, which is the 3d root of 4096. What is the fourth root of 65,536 ? The exponent corresponding to this number is 16 which divided by 4 gives for the exponent of the root 4, the number corresponding to which is 16. The answer is 16. = 2. |