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 integral`int_1^oo1/2^x dx` is convergent therefore the series `sum_(n=1)^oo 1/2^n` must also be **convergent**.