Question

Does the series `sum_(n=1)^oo((2^(n-1)3^(n+1))/n^n)` converge or diverge?

Answer 1

The series:

`sum_(n=1)^oo (2^(n-1)3^(n+1))/n^n`

is convergent.

Explanation:

Evaluate the ratio:

`abs(a_(n+1)/a_n) = (( 2^n 3^(n+2))/(n+1)^(n+1))/((2^(n-1)3^(n+1))/n^n`

`abs(a_(n+1)/a_n) = ( 2^n 3^(n+2))/ (2^(n-1)3^(n+1)) n^n / (n+1)^(n+1)`

`abs(a_(n+1)/a_n) = 6/(n+1) n^n/ (n+1)^n`

`abs(a_(n+1)/a_n) = 6/(n+1) 1/ ((n+1)/n)^n`

`abs(a_(n+1)/a_n) = 6/(n+1) 1/ (1+1/n)^n`

As:

`lim_(n->oo) 1/ (1+1/n)^n = 1/e`

we can conclude that:

`lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) 6/(n+1) 1/ (1+1/n)^n = 6/e lim_(n->oo) 1/(n+1) = 0`

and then based on the ratio test the series is convergent.