Calculus of a Single Variable, Chapter 9, 9.3, Section 9.3, Problem 30

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^(1/2) , then a_n = 1/n^(1/2) then applying a_n=f(x) , we consider:
f(x) = 1/x^(1/2) .
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^(1/2) = lim_(t-gtoo)int_1^t 1/x^(1/2) dx
Apply Law of exponent: 1/x^m = x^(-m) .
lim_(t-gtoo)int_1^t 1/x^(1/2) dx =lim_(t-gtoo)int_1^t x^(-1/2) dx
Apply Power rule for integration: int x^n dx = x^(n+1)/(n+1) .
lim_(t-gtoo)int_1^t x^(-1/2) dx=lim_(t-gtoo)[ x^(-1/2+1)/(-1/2+1)]|_1^t
=lim_(t-gtoo)[ x^(1/2)/(1/2)]|_1^t
=lim_(t-gtoo)[ x^(1/2)*(2/1)]|_1^t
=lim_(t-gtoo)[ 2x^(1/2)]|_1^t
or lim_(t-gtoo)[ 2sqrt(x)]|_1^t
Apply definite integral formula: F(x)|_a^b = F(b)-F(a) .
lim_(t-gtoo)[ 2sqrt(x)]|_1^t=lim_(t-gtoo)[2sqrt(t) -2sqrt(1)]
=lim_(t-gtoo)[2sqrt(t) -2*1]
=lim_(t-gtoo)[2sqrt(t) -2]
= oo
Note: lim_(t-gtoo)( -2) =-2 and lim_(t-gtoo)2sqrt(t) = oo then oo-2~~oo .
The lim_(t-gtoo)[ 2sqrt(x)]|_1^t = oo implies that the integral diverges.
Conclusion: The integralint_1^oo 1/x^(1/2) diverges, therefore the series sum_(n=1)^oo 1/n^(1/2) must also diverge.