1.1. The method of Chebyshev 5 The inequality ψ(x)−2ψ(x1/2) ϑ(x) ψ(x) follows since ϑ is an increasing and nonnegative function, and then log(2) lim inf x→+∞ ϑ(x) x lim sup x→+∞ ϑ(x) x 2 log(2) by our estimates on ψ(x). Clearly ϑ(x) π(x) log(x), and so π(x)/(x/ log(x)) ϑ(x)/x. Finding an upper bound for π(x) is only slightly more challenging. The inequality ϑ(x) yp≤x log(p) (π(x) π(y)) log(y) yields π(x) x/ log(y) ϑ(x) x + π(y) x/ log(y) ϑ(x) x + y x/ log(y) . To obtain the desired upper bound, we must let y increase fast enough with x so that the left-hand side of the inequality is close to π(x)/(x/ log(x)) for large x, while the second term on the right-hand side should become negligible in comparison. The choice y = x/ log2(x) works well, giving π(x) x/(log(x) 2 log log(x)) ϑ(x) x + x/ log2(x) x/(log(x) 2 log log(x)) . The second term on the right-hand side tends to zero, and π(x) x/(log(x)−2 log log(x)) π(x) x/ log(x) = log(x) 2 log log(x) log(x) 1 as x +∞. Thus lim inf x→+∞ ϑ(x) x lim inf x→+∞ π(x) x/ log(x) lim sup x→+∞ π(x) x/ log(x) lim sup x→+∞ ϑ(x) x and so log(2) lim inf x→+∞ π(x) x/ log(x) lim sup x→+∞ π(x) x/ log(x) 2 log(2). Since limx→+∞ li(x)/(x log(x)) = 1, the last inequality shows that π(x) and li(x) have the same order of growth. Choosing y = xa with 0 a 1 and a otherwise, arbitrarily yields ϑ(x) x π(x) x/ log(x) 1 a ϑ(x) x + xa x/ log(x) thus we see that π(x) x/ log(x), and π(x) li(x), and ϑ(x) x, and ψ(x) x, and a = A = 1, are equivalent formulations of the Prime Number Theorem.
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