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

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.