Prove the series converges/diverges

By comparing the partial sums of the following series with the carefully chosen integrals, prove that the series ∞ Σ n=2 of 1/(n ln n) diverges

whereas

∞ Σ n=2 of 1/(n (ln)^2 n) converges


Thanks a lot!

Using the integral test....

∞
Σ 1/(n ln n)
n=2

Let b = ∞....

lim x---> b: ∫dx / xlnx from 2 to b

u = lnx
du = dx / x

lim x---> b: ∫du / u from 2 to b

lim x---> b: ln[ln(x)] from 2 to b

Plugging in the limits gives us....

ln(∞) - ln[ln(2)] = ∞

So the integral diverges, and because the integral diverges, the series diverges as well..





∞
Σ 1/(n (ln)^2 n)
n=2

let 'b' = ∞....

lim x---> b: ∫dx / x*ln²x from 2 to b

u = lnx
du = dx / x

lim x---> b: ∫du / u² from 2 to b

Integrating....

lim x---> b: -1/ u from 2 to b

Substituting....

lim x---> b: -1 / ln[x] from 2 to b

Plugging in the limits....

-1/ln[∞] + 1/ln[2] = 0 + 1/ln[2]]

Since the Integral converges to 1/ln[2], the Series therefore converges.