`sum_(n=1)^oo 3^(-n)` Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or...

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...

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.