`sum_(n=1)^oo1/(2n+3)^3`

For the integral test, if f is positive, continuous and decreasing for `x>=1` and `a_n=f(n)` , then `sum_(n=1)^ooa_n` and `int_1^oof(x)dx` either both converge or diverge.

Now, `f(x)=1/(2x+3)^3`

Now function is positive and continuous.

Let's determine whether f(x) is decreasing, by finding its derivative `f'(x)`

`f(x)=(2x+3)^(-3)`

`f'(x)=-3(2x+3)^(-3-1)d/dx(2x+3)`

`f'(x)=-3(2x+3)^(-4)(2)`

`f'(x)=-6/(2x+3)^4`

`f'(x)<0` , so the function is decreasing

Because f(x) satisfies the conditions for the integral test, we can apply integral test.

`int_1^oo1/(2x+3)^3dx`

`=[1/2(2x+3)^(-3+1)/(-3+1)]_1^oo`

`=[1/-4(1/(2x+3)^2)]_1^oo`

`=-1/4[1/(2*oo+3)^2-1/(2*1+3)^2]`

`=-1/4[0-1/5^2]`

`=-1/4(-1/25)`

`=1/100`

So f(x) converges.

Therefore, `sum_(n=1)^oo1/(2n+3)^2` converges.

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