`sum_(n=1)^oo2/(3n+5)`

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=2/(3n+5)`

Consider `f(x)=2/(3x+5)`

Refer to the attached graph of the function. It is positive and continuous on the interval `[1,oo)`

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

`f'(x)=2(-1)(3x+5)^(-2)(3)`

`f'(x)=-6/(3x+5)^2`

So,`f'(x)<0`

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

Now let's determine whether the corresponding improper integral `int_1^oo2/(3x+5)dx` converges or diverges as:

`int_1^oo2/(3x+5)dx=lim_(b->oo)int_1^b2/(3x+5)dx`

`=lim_(b->oo)[2/3ln|3x+5|]_1^b`

`=lim_(b->oo)2/3[ln|3b+5|-ln|3+5|]`

`=2/3[oo-ln8]`

`=oo` which implies that the integral diverges.

Therefore the series `sum_(n=1)^oo2/(3n+5)` must also diverge.

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