The Integral test is applicable if f is positive and decreasing function on the infinite interval `[k, oo) ` where `kgt= 1 ` and `a_n=f(x)` . Then the series `sum_(n=1)^oo a_n` converges if and only if the improper integral `int_1^oo f(x) dx` converges. If the integral diverges then the series also diverges.

For the given series `sum_(n=1)^oo 1/n^(1/4)` , the `a_n = 1/n^(1/4)` then applying `a_n=f(x)` , we consider:

`f(x) = 1/x^(1/4)` .

As shown on the graph of f(x), the function is positive on the interval `[1,oo)` . As x at the denominator side gets larger, the function value decreases.

Therefore, we may determine the convergence of the improper integral as:

`int_1^oo 1/x^(1/4) = lim_(t-gtoo)int_1^t 1/x^(1/4) dx`

Apply the Law of exponents: `1/x^m = x^(-m)` .

`lim_(t-gtoo)int_1^t 1/x^(1/4) dx =lim_(t-gtoo)int_1^t x^(-1/4) dx`

Apply the Power rule for integration: `int x^n dx = x^(n+1)/(n+1)` .

`lim_(t-gtoo)int_1^t x^(-1/4) dx=lim_(t-gtoo)[ x^(-1/4+1)/(-1/4+1)]|_1^t`

`=lim_(t-gtoo)[ x^(3/4)/(3/4)]|_1^t`

`=lim_(t-gtoo)[ x^(3/4)*(4/3)]|_1^t`

`=lim_(t-gtoo)[ (4x^(3/4))/3]|_1^t`

Apply the definite integral formula: `F(x)|_a^b = F(b)-F(a)` .

`lim_(t-gtoo)[ (4x^(3/4))/3]|_1^t=lim_(t-gtoo)[ (4*t^(3/4))/3-(4*1^(3/4))/3]`

`=lim_(t-gtoo)[(4t^(3/4))/3-(4*1)/3]`

`=lim_(t-gtoo)[(4t^(3/4))/3-4/3]`

`= oo`

The `lim_(t-gtoo)[ (4x^(3/4))/3]|_1^t= oo` implies that the integral diverges.

Note: Divergence test states if `lim_(n-gtoo)a_n!=0 ` or does not exist then the `sum_(n=1)^oo a_n` diverges.

Conclusion: The integral `int_1^oo 1/x^(1/4) ` diverges therefore the series `sum_(n=1)^oo 1/n^(1/4)` must also diverges.