series ((-1)^n?n)/ln(n) from 1 to infinity Solution Consider the limit n^e/ln n

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Consider the limit n^e/ln n as n ? 8, where e > 0. Since both numerator and denominator approach 8, we can use l'Hopital's rule. Differentiate both numerator and denominator to obtain e n^(e-1) / (1/n) = e n^e. Clearly, as n ? 8, e n^e ? 8 as well. So in fact the asymptotic behaviour of ln n is that it is more slowly-growing than any n^e where e > 0. Still, it is clearly faster-growing than n^0 = 1, and by multiplying by n we get that n ln n is faster growing than n^(1+e) but more slowly growing than n. Hence the infinite series for 1/(n ln n) dominates the convergent series 1/(n + e) but is dominated by the divergent series 1/n, so here the comparison test fails. I hope this clears up the confusion regarding comparison with 1/n^1.000001. However, we can still use the comparison test, by making this observation concerning dominance, where lg n is the base 2 log of n: 1/(n ln n) > 1/(n lg n) > 1/(n ?lg n?) > 1/(2^?lg n? ?lg n?) Now examine the last series, for n = 3 to 8: [1/(4*2) + 1/(4*2)] + [1/(8*3) + 1/(8*3) + 1/(8*3) + 1/(8*3)] + [1/(16*4) + (6 terms) ... + 1/(16*4)] + [1/(32*5) + ... 15 terms] + ... = 2/4 * 1/2 + 4/8 * 1/3 + 8/16 * 1/4 + ... = 1/2 (1/2 + 1/3 + 1/4 + ...) which obviously diverges. and we follow the chain of dominance back up to the original series to conclude that that must diverge too. Of course, the integral test gives a shorter solution; rewrite 1/(n ln n) dn as ln n 1/n dn, and make the substitution u = ln n, then du = 1/n dn and our integral can be re-written simply du/u, from u = ln 3 to u = ln 8. The integral is then ln ln 8 - ln ln 3 which diverges, hence the original series DIVERGES.

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