Convergent and Divergent Series
Date: 07/14/98 at 20:58:52 From: Gena Subject: Convergent and Divergent Series I've tried doing these problems and reading the lessons in the textbook, however I cannot seem to solve these problems. Determine whether convergent or divergent 1 1 1 ----- + ------ + ------ + ... (1)(2) (3)(4) (5)(6) Use the comparison test 1/2 + 2/3 + 3/4 + ... Also if you could help me find more information on this topic I would really appreciate it.
Date: 07/15/98 at 10:36:37 From: Doctor Rob Subject: Re: Convergent and Divergent Series You can compare the first series, term by term, with 1/1^2 + 1/2^2 + 1/3^2 + ..., which, I hope you know, converges. The second series can be compared term by term with 1/2 + 1/3 + 1/4 + ..., which, I hope you know, diverges. With the first series, there is a trick. 1/(k*[k+1]) = 1/k - 1/(k+1), so it can be rewritten 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... As an alternating series whose terms go to zero, it must converge to something. It turns out that that something is ln(2). As far as general advice, the comparison test can be used if lim a(n)*n > L > 0 to show divergence, by comparing with the harmonic series multiplied by L, as above. Similarly, it can be used if lim a(n)*n^[1+e] < L < infinity with e > 0, to show convergence, by comparing with the series Sum L/n^(1+e), which converges if e > 0. It cannot be used in conjunction with a series of the form a(n) = 1/n^(1+e) if lim a(n)*n = 0 but lim a(n)*n^(1+e) = infinity for any e > 0. (Example: Sum 1/[n*log(n)].) It might be used with some other series known to converge or diverge. It may or may not be used if the indicated limits do not exist. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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