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.
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