11. Given the infinite series 2 M ((v+2); +3(v+2)3+ 5(v+2)3 +&c.) to find the limit of convergency. Here the general convergency is v2 (2-3) : therefore the li 2xv2 mit is 2x(v+2) This series will therefore always con v2 verge, whatever be the value of v; but the greater v is, the less wili be the convergency. This and the two preceding series have each the logarithm of v+1 for their value. The first converges the least, the ultimate rate being v; the second converges something more than the first, the ultimate rate being the last converges the most, the ultimate rate being In the first, v must be a fraction; in the second and last examples the series will always converge, whatever be the value of v. 12. Given the infinite series 2 ((2a+v) c.) to +&c. to find its limit of convergency. Here the general convergency is ༧• ·+ 3 (2a+v)3 +5(2a+v)° The value of this series added to the logarithm of a, gives the logarithm of a+v. The convergency may be increased in any degree, according as a is greater than v. N.B. As the series in the last example may be derived from this by making a=1, so may also the limit of convergency be derived from the limit here found, and thus the rate of convergency will be lessened. 13. Find the limit of convergency of the infinite series (1+v)+% 21% (1+v)2+231*(1+2)3 + 241%*(1+v')* 1.2.323v3 + &c. Here the general convergency is (T+v) [z+x(x−1)]' ย 1+v is the limit. When v is very great, approaches to unity, 1+v but never will equal it. If v be a fraction, the convergency will be greater, the greater the denominator is. This series is a transformation of p—qv+rv2—sv3+ &c. and is particularly useful in finding a rapid approximation to such slow con both cases, v would be equal to unity, and 2 the first term of the series, z=1 in the first, and 2 in the second. vnx บ =nar a Hence the convergency will be rapid when v is small compared with a: when x=3, the convergency in the third term is v(n-m) 2na The series here proposed is the expansion of (a+v). Therefore to find the arithmetical expansion of (a+v), a ought to be a large square number, and v small. Thus, to find the square root of 5; then m=1, m n=2, make a =4, v=1, then a=4=2; the above series then be comes, 3)2343 8) 781 97 1.11803 2 2 23606, which is the square root of 5 true to the last figure. And thus any power or root may be extracted. The ultimate convergency depending on the numbers v and a, which is here, and not upon the indices; the series will, however, converge slower, the higher the root is to be extracted. 5 4)485 8)121 L 15 7 -5)105 8) 21 Problem 3. A series being given to find the value of a, the first factor, so that it may obtain a given convergency at the rth term. Find the convergency of the series, and for z substitute its value in terms of x, and the first term a; then make the formula, thus altered, equal to the given convergency, and the value of a will be found by this equation. ue of a, so that the series may obtain a given convergency term of the series. ther term is x(x-1)+a; which, being substituted in the general a+x(x-2) a+x(x-2) ; then the resolu a+s (m+x—2) = — gives a= + Here x=2, m=5, x=4, b=3; therefore 2(5+4-2)-2.3 (4-2) 3-1 3.5.7.9.11 1 1.3.5.7.9 + &c. will obtain a convergency of in the fourth term. find the value of a, so as to make the convergency in the rth Two series which do not converge at the same rate being given, to find the number of the term, if possible, in which their convergencies are equal. Find the convergency of each series; then make the two expres sions equal; and the value of x being found, that of r will be ob tained from the equation z=z(x-1+6. Or, find the convergency of each series in terms of r and a, the first factor, and make the two expressions equal; then the value of x in the equation will give the number of the term. (a+4)R+B. 2.4 3.4 C(a+12) R + &c. being given, to find the number of the term when their convergencies are equal. The convergency of the first of these series is 4x-8+a 9[4(x-1)+a] R[4(x−1)+a]; therefore, putting these the term, as required. Let a=33; then will x=59. |