Testing ∑1/(n·ln(n))

Real Analysis · Axiom Academy

EXAMPLE Testing with the Integral Test Apply the integral test, and see why terms shrinking to zero is not enough to make a series converge. Determine whether the series converges or diverges. (The sum starts at n = 2 because would make the first term undefined.) Use the integral test. The picture behind the substitution The integral test compares the series to the area under for . Substituting turns that area into the area under — and the area under 1/u grows without bound. Because the area under 1/u is infinite, the integral diverges — and so does the series. Nice work. You applied the integral test from end to end and met a counterintuitive fact about convergence. The hypotheses matter: the integral test applies only because is positive, continuous, and decreasing on . The substitution is the whole trick: matches the 1/x already in the integrand, collapsing into . The integral diverges: , so by the integral test the series diverges. Shrinking terms are not enough: here , yet the series still diverges. The n th-term test can prove divergence, never convergence. The log barely fails: it is a hair past the harmonic series . In fact converges if and only if , and here p=1 . The integral test turns a discrete sum into a continuous area, where a substitution does the hard work — and it reveals just how slowly a series can diverge.

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