Integral test for series $\sum_{n = 18}^{\infty} \frac{n^2}{(n^3 + 3)^{7/2}}$

Question

I am stuck on how to more so algebraically to solve this problem. I understand that you would rewrite the series as a function of x, and then evaluate the integral from 18 to infinity - but that's all I got. Any pointers? Thank you in advance.

Use the integral test to determine whether the infinite series is convergent. $$\sum_{n = 18}^{\infty} \frac{n^2}{(n^3 + 3)^{7/2}}.$$

Answer 1

$\displaystyle\int_{18}^{\infty}\dfrac{x^{2}}{(x^{3}+3)^{7/2}}dx=\dfrac{1}{3}\dfrac{1}{(-7/2)+1}(x^{3}+2)^{-(7/2)+1}\bigg|_{x=18}^{x=\infty}<\infty$.

One needs to check the monotonicity for $x\rightarrow\dfrac{x^{2}}{(x^{3}+3)^{7/2}}$ on $[18,\infty)$.

Answer 2

The integral test states that if the improper integral converges then the sum converges. It is probably nearly impossible to determine what the sum is exactly, but with the integral test, you can show that it is a finite number. The improper integral can be solved using a $u$-substitution of $u=x^3+3$.

Answer 3

Let $$u=x^3+3$$ to evaluate the $$\int_{18}^{\infty} \frac{x^2}{(x^3 + 3)^{7/2}} dx$$