Question

is the Σ (n^n/(2^n)(n!)) from 0 to infinity convergent

Answer 1

Consider the ratio of the (n+1)th element in the series to the nth:\[u _{n+1}/u _{n}=((n+1)^{n+1}/(n+1)!2^{n+1}) \times (n! 2^{n}/n ^{n}\]This simplifies to \[(n+1)^{n+1}/(n ^{n}2(n+1))\]which can be written as \[((n+1)/n)^{n} \times (n+1)/(2(n+1))\]which gives \[(1/2) \times(1+1/n)^{n}\]As n tends to infinity this tends to \[e/2\] which is greater than 1. Hence, by the Ratio Test, the infinite series diverges.

Answer 2

thanks

Answer 3

I noticed my mistake I forgot to do (n+1) as the denominator, I still left it as n.Thanks