Some series of interest in probability.

∑_n≥1 1/n   diverges:

∑_n≥1 1/n > ₁∫^∞ 1/x = [log x]_1..∞ = ∞
So pr(n) ∝ 1/n cannot be a proper probability distribution on the positive integers, {1, 2, 3, ...}.

∑_i≥1 1/(n(n+1))   converges to one:

1/(n(n+1)) = 1/n - 1/(n+1),
so ∑_i≥1 1/(n(n+1)) = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ...  = 1

∑_n≥2 1/(n log_e(n))   diverges:

^d/_dx log_e(log_e(x)) = 1/(x log(x)) and,
in the integral test
₂∫^∞ 1/(x log_e(x)) = [log_e(log_e(x))]_2..∞ = ∞

Similarly   ∑_n≥2 1/{n log_e(n) log_e(log_e(n))}   diverges

because ^d/_dx log_e(log_e(log_e(x))) = 1/(x log_e(x) log_e(log_e(x)))

etc.

∑_n≥2 1/(n (log(n))²)   converges:

^d/_dx 1/log_e(x) = – 1/(x (log_e(x))²)
and [-1/log(x)]_2..∞ = 1/log(2) is finite.

Similarly   ∑_n≥2 1/(n (log(n))^p)   converges provided p>1.0,

see above.