sum_(n=1)^oo 1/(2n+3)^3 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.

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.