`sum_(n=1)^oo(4n)/(2n+1)`

The integral test is applicable if f is positive, continuous and decreasing function on infinite interval `[k,oo)` where `k>=1` and `a_n=f(x)` . Then the series `sum_(n=1)^ooa_n` converges or diverges if and only if the improper integral `int_1^oof(x)dx` converges or diverges.

For the given series `a_n=(4n)/(2n+1)`

Consider `f(x)=(4x)/(2x+1)`

Refer to the attached...

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`sum_(n=1)^oo(4n)/(2n+1)`

The integral test is applicable if f is positive, continuous and decreasing function on infinite interval `[k,oo)` where `k>=1` and `a_n=f(x)` . Then the series `sum_(n=1)^ooa_n` converges or diverges if and only if the improper integral `int_1^oof(x)dx` converges or diverges.

For the given series `a_n=(4n)/(2n+1)`

Consider `f(x)=(4x)/(2x+1)`

Refer to the attached graph of the function. From the graph we observe that the function is positive and continuous. However it is not decreasing on the interval `[1,oo)`

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

Let's find the derivative by the quotient rule:

`f(x)=(4x)/(2x+1)`

`f'(x)=((2x+1)d/dx(4x)-(4x)d/dx(2x+1))/(2x+1)^2`

`f'(x)=((2x+1)(4)-(4x)(2))/(2x+1)^2`

`f'(x)=(8x+4-8x)/(2x+1)^2`

`f'(x)=4/(2x+1)^2`

So `f'(x)>0`

which implies that the function is not decreasing.

Since the function does not satisfies the conditions for the integral test, we can not apply integral test.