consider S(a,b) = ?_(k=0)^? 1/((k+a)(k+b)) a) show that S(a,b) converges if a an
ID: 3190835 • Letter: C
Question
consider S(a,b) = ?_(k=0)^? 1/((k+a)(k+b)) a) show that S(a,b) converges if a and b are both positive. Why cant either be negative? b) find a general formula for S(a, a+m) using telescoping. c) find S(2,6)Explanation / Answer
It can still converge for a or b negative as long as the negative value is not an integer. If it is an integer, however, then when k = -a or -b, there is a 0 in the denominator. b) The general formula is clearly meant for the case where m is a positive integer Then 1/((k+a)(k+a+m)) = 1/m (1/k+a - 1/(k+a+m)) Then, as we sum these values, for all k >= m, we will be adding the value on the left side that we have previously subtracted on the right side. (1/k+a on the left = 1/(k-m + a + m)) Thus, the unique terms that we add on the left are 1/m(1/a+1/a+1...+1/a+m-1) c)S(2,6) = S(2,2+4) (a = 2; m = 4) Thus, the sum is 1/4(1/2+1/3+1/4+1/5) = 1/4(30+20+15+12)/60 = 77/240
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.