Question

lim (1!+2!+3!+...+n!)/(2n)! .. is ok with Stolz–Cesàro theorem? = lim (n+1)!/(2n+2)!-(2n)!..

Answer 1

yes, I think that's right, but is the new limit easier to calculate than the first?

Answer 2

let a_n = (1! + 2! + ... + n!)/(2n!). Then clearly 0 < a_n for all n.

Now \[ 0 < a_n < n . n!/(2n!) \]
Keep on bounding this and show the upper bound also goes to zero.