Calculus of a Single Variable, Chapter 9, 9.3, Section 9.3, Problem 29

Recall that 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^3 , the a_n = 1/n^3 then applying a_n=f(x) , we consider:
f(x) = 1/x^3 . The function is positive and 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^3 = lim_(t-gtoo)int_1^t 1/x^3 dx
Apply the Law of exponent: 1/x^m = x^(-m) .
lim_(t-gtoo)int_1^t 1/x^3 dx =lim_(t-gtoo)int_1^t x^(-3) dx
Apply Power rule for integration: int x^n dx = x^(n+1)/(n+1).
lim_(t-gtoo)int_1^t 1/x^3 dx =lim_(t-gtoo)[ x^(-3+1)/(-3+1)]|_1^t
=lim_(t-gtoo)[ x^(-2)/(-2)]|_1^t
=lim_(t-gtoo)[ -1/(2x^2)]|_1^t
Apply the definite integral formula: F(x)|_a^b = F(b)-F(a) .
lim_(t-gtoo)[ -1/(2x^2)]|_1^t=lim_(t-gtoo)[-1/(2*t^2) -(-1/(2*1^2))]
=lim_(t-gtoo)[ -1/(2t^2)-(-1/2)]
=lim_(t-gtoo)[-1/(2t^2)+1/2]
= 1/2 .
Note: lim_(t-gtoo) 1/2 =1/2 and lim_(t-gtoo)1/(2t^2) = 1/oo or 0
The integral int_1^oo 1/x^3 is convergent therefore the series sum_(n=1)^oo 1/n^3 must also be convergent.