Calculus Help: Convergent and Divergent: Summation from k=1 to infinity 3^k/(2k)^k or 3^n/(2n)^n

Calculus Help: Convergent and Divergent: Summation from k=1 to infinity 3^k/(2k)^k or 3^n/(2n)^n

Join this channel to get access to perks:    / @calculusphysicschemaccountingt   Here is the technique to solve this question related to convergence and divergence. #Techniques #Solutions #FOrmula