MAT157Y1 Lecture : Series Convergent.pdf
Document Summary
Find the values of p for which the series is convergent: We want to compare this with f(x) = that for p = 0 the series becomes. 1 and it diverges, so we can assume that p 6= 0. 1 negative for big enough x. f0(x) = 0 ((ln x)p+xp(ln x)p 1 1 x ) x2(ln x)2p. The numerator will be negative whenever lnx > 0 (ln x + p) > 0, which means x > 1 and ln x > p x > e p. Thus, f0 is negative for x big enough and f will be decreasing. This is what i think i explained rather confusingly in class: it only matters that f is eventually decreasing, because that corresponds to the tail" of the series where we are adding in nitely many terms. 1 x(ln x)p dx, p 6= 0 to determine if the series is convergent (even with the 2, not e p as the lower limit).
