Wyzant T.
asked • 05/03/24
Yefim S. answered • 05/03/24
We can use limit comparison test: compare with series ∑n=1∞1/n3/2 which converges
Then lim n→∞(lnn/n2/1/n3/2) = lim n→∞(lnn/n1/2) = 0 by Llopital rule
So given series converges also.
Doug C. answered • 05/04/24
desmos.com/calculator/hlrkpe1nia
Doug C.
Addition to the board with limit comparison test. If limit equals zero, then if sum(b_n) converges, so does sum (a_n). Since b_n was a p-series, which converges, so does a_n. So, limit comparison is much easier to use than integral test.05/05/24
Mark M. answered • 05/03/24
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
= limb→∞ [-lnb / b + ln2/2] + limb→∞ [∫(2 to b) (1 / x2)dx]
= 0 + ln2/2 + limb→∞ [ -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.
Wyzant T.
could you go through that please? I dont fully understand the Integral test.05/03/24
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Doug C.
For limit comparison, the limit must be greater than 0 (and less than infinity). I do not think that test applies?05/04/24