Series ConvergenceDate: 01/27/2001 at 12:39:50 From: Chris Subject: Series Convergence My calculus teacher gave me a review sheet with sample test questions. The questions on series convergences are really tough! If you can help me out with any of them, that would be a great help. Test each of the following series for convergence. If the series is alternating, tell whether the convergence is conditional or absolute: a) infinity E ((-1)^(k+2)) ((3k + 2) / ((2k + 3)^2)) k = 1 Clarification: -1 is raised to the power k+2; 3k + 2 is divided by (2k + 3)^2 b) infinity E ((-1)^(k)) (2k)! (e/2)^k (2^(3k)) (k!)^2 k = 0 Clarification: -1 is raised to the power k; 2k factorial is divided by 2 raised to the power 3k, times k factorial squared; this is all multiplied by e divided by 2, raised to the power k. c) infinity E ((-1)^k) ((1/square root of k) - (1/square root of k + 5)) k = 1 Clarification: -1 is raised to the k; k and k+5 are under square root signs, and both are in denominator spots. Date: 01/28/2001 at 11:29:06 From: Doctor Fenton Subject: Re: Series Convergence Hi Chris, Thanks for writing to Dr. Math. When I examine series like these, I like to simplify the terms to a point where I can "see" what is happening. In your first problem, although it is an alternating series, you should first ignore the +/- 1 produced by the power of (-1) and check absolute convergence. The key idea is that for very large k, only the largest power of k in each term will be significant: 3k+2 is essentially 3k, and (2k+3)^2 is essentially (2k)^2 = 4k^2. For very large k, then, the terms are essentially 3k 3 ------ = ---- . 4k^2 4k If you want to be more formal, write 3k+2 3k (1+[2/(3k)]) --------- = ----------------------- (2k+3)^2 (2k)^2 * (1+[3/(2k)])^2 The second factor in the numerator and denominator is clearly -> 1 as k->oo . So, essentially, the first problem is like a p-series with p = 1 (the p-series is the sum of the reciprocal p-th powers of the positive integers), so this is like the harmonic series (or better, 3/4 times the harmonic series), and you should expect absolute divergence and conditional convergence of the alternating series. If you have the Limit Comparison Test, you can use this; otherwise you need to make a direct comparison with, say, (1/2) times the harmonic series to show absolute divergence. To apply the Alternating Series Test, you need to show that the terms alternate in sign (given), and that the terms decrease to 0. The expression above shows that the limit is 0, but you need to show that the formula for the terms is decreasing. Try showing that f'(x)<0 for f(x)=(3k+2)/(2k+3)^2 . The second series, with factorials and exponentials, is tailor-made for the Ratio Test. If that doesn't work, you've got a really difficult series (but in your case, it WILL work, and it gives you the complete answer). For the third problem, it's difficult to see what's happening because the terms consist of small numbers being subtracted. Rewrite them as 1 1 sqrt(k+5) - sqrt(k) ------- - ---------- = --------------------- sqrt(k) sqrt(k+5) sqrt( k*(k+5)) but that's still unclear, since there's still a subtraction in the numerator, so rationalize the numerator: sqrt(k+5) - sqrt(k) sqrt(k+5) + sqrt(k) -------------------- * ------------------- sqrt( k*(k+5)) sqrt(k+5) + sqrt(k) (k+5) - k = -------------------------------------- sqrt( k*(k+5)) [sqrt(k+5) + sqrt(k)] 5 = -------------------------------------- sqrt( k*(k+5)) [sqrt(k+5) + sqrt(k)] Again, for large k, the first term is sqrt(k(k+5)) ~ k, and the second term is a sum of two terms, each of which is like sqrt(k), so the series is like 5 -------- k^(3/2) which is another p-series, whose behavior you should know. If you have further questions, please write again. - Doctor Fenton, The Math Forum http://mathforum.org/dr.math/ |
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