sum_(n=1)^oo 1/(nroot(4)(n)) Determine the convergence or divergence of the series.

To evaluate the given series: sum_(n=1)^oo 1/(nroot(4)n) , we may apply  radical property: root(a)(x^b) = x^(b/a) and Law of exponent: x^a*x^b= x^(a+b) .
The given series becomes :
sum_(n=1)^oo 1/(nroot(4)n) =sum_(n=1)^oo 1/(n*n^(1/4))
                    =sum_(n=1)^oo 1/n^(1/4+1)
                     =sum_(n=1)^oo 1/n^(5/4)
                     or  sum_(n=1)^oo 1/n^(1.25)
 The sum_(n=1)^oo 1/n^(1.25) is in a form of a p-series.
In general, the p-series follows the following form:
sum_(n=1)^oo 1/n^p =1/1^p + 1/2^p+ 1/3^p +1/4^p + 1/5^p +...
Recall the theorem for a p-series states that sum_(n=1)^oo 1/n^p is convergent if pgt1 and divergent if plt=1 .
Applying the theorem on the given series sum_(n=1)^oo 1/n^(1.25) , we compare n^(1.25) with n^p to determine the corresponding value: p =1.25 .
It satisfies pgt1 since 1.25gt1 .
Therefore, the given series sum_(n=1)^oo 1/(nroot(4)n) is convergent.