`1/2 + 1/5 + 1/10+1/17+1/26+...` 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

`1/2+1/5+1/10+1/17+1/26+............`

The series can be written as,

`1/(1^2+1)+1/(2^2+1)+1/(3^2+1)+1/(4^2+1)+1/(5^2+1)+..........`

So based on the above pattern we can write the series as,

`sum_(n=1)^oo1/(n^2+1)`

The integral test is applicable if f is positive, continuous and decreasing function on the infinite interval `[k,oo)` where `k>=1` and `a_n=f(x)` . Then the series converges or diverges...

See
This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Get 48 Hours Free Access

`1/2+1/5+1/10+1/17+1/26+............`

The series can be written as,

`1/(1^2+1)+1/(2^2+1)+1/(3^2+1)+1/(4^2+1)+1/(5^2+1)+..........`

So based on the above pattern we can write the series as,

`sum_(n=1)^oo1/(n^2+1)`

The integral test is applicable if f is positive, continuous and decreasing function on the infinite interval `[k,oo)` where `k>=1` and `a_n=f(x)` . Then the series converges or diverges if and only if the improper integral `int_k^oof(x)dx` converges or diverges.

For the given series `a_n=1/(n^2+1)`

Consider `f(x)=1/(x^2+1)`

 

Graph of the function is attached. From the graph we can see that the function is positive, continuous and decreasing on the interval `[1,oo)`

Since the function satisfies the conditions for the integral test, we can apply integral test.

Now let's determine whether the corresponding improper integral `int_1^oo1/(x^2+1)dx` converges or diverges.

`int_1^oo1/(x^2+1)dx=lim_(b->oo)int_1^b1/(x^2+1)dx`

Use the common integral: `int1/(x^2+1)dx=arctan(x)`

`=lim_(b->oo)[arctan(x)]_1^b`

`=lim_(b->oo)[arctan(b)-arctan(1)]`

`=lim_(b->oo)arctan(b)-arctan(1)`

`=pi/2-pi/4`

`=pi/4`

Since the integral `int_1^oo1/(x^2+1)dx` converges, we conclude from the integral test that the series also converges.

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)
Approved by eNotes Editorial Team