Student Question

`sum_(n=1)^oo 1/2^n` Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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.

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial