Calculus of a Single Variable, Chapter 9, 9.3, Section 9.3, Problem 4

Integral test is applicable if f is positive and decreasing function on interval [k,oo) where a_n = f(x) .
If the integral int_k^oo f(x) dx is convergent then the series sum_(n=k)^oo a_n is also convergent.
If the integral int_k^oo f(x) dx is divergent then the series sum_(n=k)^oo a_n is also divergent.
For the series sum_(n=1)^oo 3^(-n) , we have a_n=3^(-n) then we may let the function:
f(x) = 3^(-x) which has the below graph:

As shown on the graph, f(x) is positive and decreasing on the interval [1,oo) . This confirms that we may apply the Integral test to determine the convergence or divergence of a series as:
int_1^oo 3^(-x) dx =lim_(t-gtoo)int_1^t 3^(-x)dx
To determine the indefinite integral of int_1^t 3^(-x)dx , we may apply u-substitution by letting: u =-x then du = -dx or -1du =dx .
The integral becomes:
int 3^(-x) dx =int 3^u * -1 du
= - int 3^u du
Apply the integration formula for an exponential function: int a^u du = a^u/ln(a) +C where a is a constant.
- int 3^u du =- 3^u/ln(2)
Plugging-in u =-x on - 3^u/ln(3) , we get:
int_1^t 3^(-x)dx= -3^(-x)/ln(3)|_1^t
= - 1/(3^xln(3))|_1^t
Applying the definite integral formula: F(x)|_a^b = F(b)-F(a) .
- 1/(3^xln(3))|_1^t= [- 1/(3^tln(3))] - [- 1/(3^1ln(3))]
=- 1/(3^tln(3)) + 1/(3ln(3))
=- 1/(3^tln(3)) + 1/ln(27)
Note: 3 ln(3)= ln(3^3) = ln(27)
Apply int_1^t 3^(-x) dx=- 1/(3^tln(3)) + 1/ln(27) , we get:
lim_(t-gtoo)int_1^t 3^(-x) dx=lim_(t-gtoo)[- 1/(3^tln(3)) + 1/ln(27)]
=lim_(t-gtoo)- 1/(3^tln(3)) +lim_(t-gtoo) 1/ln(27)
= 0 +1/ln(27)
=1/ln(27)
Note: 3^ooln(3) =oo then 1/oo =0 .
The lim_(t-gtoo)int_1^t 3^(-x)dx=1/ln(27) implies the integral converges.
Conclusion:
The integral int_1^oo 3^(-x)dx is convergent therefore the series sum_(n=1)^oo 3^(-n) must also be convergent.