Question

sum of 3^(1/n) from 1 to infinity

Answer 1

really confused on how to do this would you use ratio test, comparisnt test?

Answer 2

anyone?

Answer 3

the series looks like this:
\(3+3^{1/2}+3^{1/3}+\ldots\)
an exponent of (1/n) implies n^th root
and n^th root of "3" will always be greater than "1"
and hence the sum will keep diverging.
\[
L=\lim_{n\to\infty}3^{1/n}\\
\text{substitute}\quad u=1/n\qquad\therefore\text{as}\;n\to\infty,u\to0\\
L=\lim_{u\to0}3^u=1\ne0
\]
by Limit test, the series diverges@@