11.3) Determine whether the series is convergent or divergent.

Question

Answer 1

We can use limit comparison test: compare with series ∑_n=1^∞1/n^3/2 which converges

Then lim n→∞(lnn/n²/1/n^3/2) = lim n→∞(lnn/n^1/2) = 0 by Llopital rule

So given series converges also.

Answer 2

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Answer 3

Use the integral test. Integrate by parts with u = lnx and dv = (1/x²)dx. Then du = (1/x)dx and v = -1/x.

∫_{(2 to ∞)} [lnx / x²]dx = (-lnx / x)_{(2 to ∞)} + ∫_{(2 to ∞)} [1 / x²]dx

= lim_b→∞ [-lnb / b + ln2/2] + lim_b→∞ [∫_{(2 to b)} (1 / x²)dx]

= 0 + ln2/2 + lim_b→∞ [ -1/x_{(2 to b)} ] = ln2 / 2 - 0 + 1/2

The series has all positive terms, f(x) = lnx / x² is decreasing and continuous for x≥2, and the integral of f(x) from 2 to infinity converges.

So, by the integral test, the series converges.