I need to test for converges for a few series. Can someone help. Please explain so I can follow reasoning.?

Question

Sum of 1/ln k k=2 -> infinity

Sum 1/k! k=1 -> infinity

sum ln k/k^2

Answer 1

Sorry for just a partial answer at this point...

The second one converges by the ratio test.

[ 1/(k+1)! / 1/k! ] = [ k! / (k+1)! ] = 1/(k+1) -> 0 as k->infinity

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Edit #1: The first series is divergent by the Comparison Test. Compare the given series with the divergent series ∑ 1/k.

1/ln(k) > 1/k for all k ≥ 2.

{ The Comparison Test says that if you can show that all of a given series terms are greater than (or equal to) the corresponding terms of a known divergent series, then the given series is also divergent. }

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Edit #2: Third series convergent by Integral Test.

∫ ln(x) / x^2 dx [by parts]

Let u = ln(x), then du = (1/x) dx

Let dv = 1/(x^2) dx. then v = -1/x

∫ ln(x) / x^2 dx

= -ln(x)/x + ∫ 1/(x^2) dx

= -ln(x)/x - 1/x

= - (ln(x) + 1) / x

To determine if this integral converges, we need to look at

limit as t->infinity ∫ ln(x) / x^2 dx for x=2 to x=t

= limit as t->infinity [ - (ln(x) + 1) / x ] for x=2 to x=t

= limit as t->infinity [ - (ln(t) + 1) / t  ] - [ - (ln(2) + 1) / 2 ]

= (ln(2) + 1) / 2.

The improper integral  Ã¢ÂˆÂ« ln(x) / x^2 dx is convergent, therefore the original series is convergent.

{ Note: limit as t->infinity [ - (ln(t) + 1) / t  ] = 0 by L'Hopital's rule }