Question

Question #c229d

Answer 1

By using the Stirling's asymptotic approximation formula

https://en.wikipedia.org/wiki/Stirling%27s_approximation

`a_n = (e/n)^n approx sqrt(2pi n)/(n!)` and then

`sum_(k=1)^oo a_n le sqrt(2pi)sum_(k=1)^oo n/(n!) = sqrt(2pi)sum_(k=1)^oo 1/((k-1)!) = sqrt(2pi) e`

so it converges.

now considering

`sum_(k=2)^oo 1/(log k)^(log k)` making `logk = y`

`sum_(k=2)^oo 1/y^y =sum_(k=2)^oo (1/y)^y lt sum_(k=2)^oo (e/y)^y` hence, according with the previous considerations,

`sum_(k=2)^oo 1/(log k)^(log k)`

converges.