`1/3+1/5+1/7+1/9+1/11+..........`

The series can be written as,

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

Based on the above pattern we can write the series as,

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

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

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`1/3+1/5+1/7+1/9+1/11+..........`

The series can be written as,

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

Based on the above pattern we can write the series as,

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

The integral test is applicable if f is positive, continuous and decreasing function on the 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/(2n+1)`

Consider `f(x)=1/(2x+1)`

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

We can also determine whether function is decreasing by finding the derivative f'(x) such that `f'(x)<0` for `x>=1`

We can apply the integral test, as the function satisfies the conditions for the integral test.

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

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

Let's first evaluate the indefinite integral `int1/(2x+1)dx`

Apply integral substitution:`u=2x+1`

`=>du=2dx`

`int1/(2x+1)dx=int1/u(du)/2`

Take the constant out and use common integral:`int1/xdx=ln|x|`

`=1/2ln|u|`

Substitute back `u=2x+1`

`=1/2ln|2x+1|+C` where C is a constant

`int_1^oo1/(2x+1)dx=lim_(b->oo)[1/2ln|2x+1|]_1^b`

`=lim_(b->oo)1/2[ln|2b+1|-ln|2(1)+1|]`

`=oo-ln3/2`

`=oo`

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