The way that the integral test is taught is that
If f is a continuous, positive and decreasing function where $f(n)=a_n$ on the interval [1,∞), then. the improper integral $\int_1^\infty f(x)dx$ and the infinite series $\sum_1^\infty a_n$. either both converge or both diverge.
However I am having quite a hard time building an intuition for this, say we have the infinite sum: $$\sum^\infty_{n=1}\frac{\ln(n)}{n^3}$$ And we compare it to the function $$f(x)=\frac{\ln(x)}{x^3}$$
At the upper bound: $$\sum^\infty_{n=1}\frac{\ln(n)}{n^3}< \int^n_1\frac{\ln(x)}{x^3}dx$$ Solving by parts $$\sum^\infty_{n=1}\frac{\ln(n)}{n^3}< \left[\frac{-\ln(x)}{2x^2} \right]^n_1 +\int^n_1\frac{1}{2x^3}$$ $$\sum^\infty_{n=1}\frac{\ln(n)}{n^3}< \frac{-\ln(n)}{2n^2} +\left[\frac{-1}{4x^2}\right]^n_1$$ $$\sum^\infty_{n=1}\frac{\ln(n)}{n^3}< \frac{-\ln(n)}{2n^2} -\frac{1}{4n^2} +\frac{1}{4}$$
Then by using L'Hôpital's rule
$$\lim \limits_{n \to \infty} \frac{-\ln(n)}{2n^2}=\lim \limits_{n \to \infty}\frac{-1}{4n^2}=0$$
Hence the integral converges to $\frac{1}{4}$ and so the sum also converges.
However looking at the graph of the function and by differentiating it it is apparent that the function only retains positive and decreasing for the interval for $x>e^{\frac{1}{3}}$
Yet the integral test seems to be still working properly over the greater interval of $x>1$ which contains both the increasing and decreasing parts of the function.
By drawing rectangles of unit width touching the function at the right corners it becomes apparent that for values $x>e^{\frac{1}{3}}$ the rectangles estimate the upper bound but for values less than this interval from $e^{\frac{1}{3}}>x>1$ they instead approximate the lower bound.
Why then does this test still apparently work and would this extend from $x=0$ to $x=\infty$ as well?
A series $\sum_{n=1}^\infty a_n$ converges if and only if, for some $N\in\Bbb N$, the series $\sum_{n=N}^\infty a_n$ converges. In other words, the first terms of the series do not matter when we are dealing with the proeblem of deciding whether a series converges or not. So, when we use the integral test, all that matters about $f$ is that, for some $a>0$, its restriction to $[a,\infty)$ is positive, decreasing and convergent to $0$. What happens to the left of that $a$ does not matter.