Test for ConvergenceDate: 8/20/96 at 17:28:15 From: Anonymous Subject: Test for Convergence I cannot solve this problem: Test the following series for convergence. sum {from k=0 to infinity} [{log(k+1)-log k}/tan^(-1) (2/k)] Tan^(-1) is tangent inverse. Because -1 < tan^(-1) < 1, I used the comparison test, but I did not achieve the goal. Thanks for your kindness. Date: 8/21/96 at 15:49:36 From: Doctor Anthony Subject: Test for Convergence This series appears to diverge since the term when k -> infinity is not zero. When k->inf term is ln(1+1/k)/tan^(-1)(2/k) As this tends to 0/0 we can use l'Hopital's rule to find its limiting value. Differentiating top and bottom separately we get: {1/(1+1/k)}(-1/k^2) ------------------- {1/(1+4/k^2)}(-2/k^2) = (1/2){k/(k+1)} -------------- k^2/(k^2+4) = (1/2)k(k^2+4) ------------- k^2(k+1) = (1/2)(k^2+4) ------------ k^2 + k = (1/2)(1+4/k^2) -------------- 1 + 1/k If we let k->inf. This ratio -> 1/2 We can only get convergence if the terms tend to zero as k-> infinity, and this is not happening, so the series diverges. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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