Calculus of a Single Variable, Chapter 9, 9.3, Section 9.3, Problem 3

Recall the Integral test is applicable if f is positive and decreasing function on interval [k,oo) where kgt=1 and a_n = f(x) .
If int_k^oo f(x) dx is convergent then the series sum_(n=k)^oo a_n is also convergent.
If 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 1/2^n , we have a_n=1/2^n then we may let the function:
f(x) = 1/2^x with a graph of:


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 converge or divergence of a series as:
int_1^oo 1/2^x dx =lim_(t-gtoo)int_1^t 1/2^x dx
To evaluate the integral of int_1^t 1/2^x dx , we may Law of exponent: 1/x^n = x^(-n) .
int_1^t 1/2^x dx =int_1^t 2^(-x) dx
To determine the indefinite integral of int_1^t 2^(-x) dx , we may apply u-substitution by letting: u =-x then du = -dx or -1du =dx .
The integral becomes:
int 2^(-x) dx =int 2^u * -1 du
= - int 2^u du
Apply integration formula for exponential function: int a^u du = a^u/ln(a) +C where a is constant.
- int 2^u du =- 2^u/ln(2)
Plug-in u =-x on - 2^u/ln(2) , we get:
int_1^t 1/2^x dx= -2^(-x)/ln(2)|_1^t
= - 1/(2^xln(2))|_1^t
Applying definite integral formula: F(x)|_a^b = F(b)-F(a) .
- 1/(2^xln(2))|_1^t = [- 1/(2^tln(2))] - [- 1/(2^1ln(2))]
=- 1/(2^tln(2)) + 1/(2ln(2))
=- 1/(2^tln(2)) + 1/ln(4)
Note: 2 ln(2)= ln(2^2) = ln(4)
Apply int_1^t 1/2^x dx=- 1/(2^tln(2)) + 1/ln(4) , we get:
lim_(t-gtoo)int_1^t 1/2^x dx=lim_(t-gtoo)[- 1/(2^tln(2)) + 1/ln(4)]
=lim_(t-gtoo)- 1/(2^tln(2)) +lim_(t-gtoo) 1/ln(4)
= 0 +1/ln(4)
=1/ln(4)

Note: 2^oo =oo and oo*ln(2) =oo then 1/oo = 0
The lim_(t-gtoo)int_1^t 1/2^x=1/ln(4) implies the integral converges.
Conclusion:

The integralint_1^oo1/2^x dx is convergent therefore the series sum_(n=1)^oo 1/2^n must also be convergent.