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

Question

1233 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-15T22:25:12

The series:

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

is convergent.

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.

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