# Is the convergence test for integrals limited to series and improper integrals?

In my Calculus text book there is a theorem named The Integral Test that states:

If the function $$f$$ is continous, positive and nondecreasing on the interval $$x\in[a, \infty)$$, then

$$\sum^{\infty}_{x=1} f(x) \text{ converges} \Leftrightarrow \int^{\infty}_{a} f(x) \text{ converges}$$

Based on my home made proof, the more general result

$$\sum^{b}_{x=1} f(x) \text{ converges} \Leftrightarrow \int^{b}_{a} f(x) \text{ converges}$$

should also hold.

Is my result true or is indeed The Integral Test limited to series and improper integrals?

• Can you come up with a finite sum, or a finite integral over a continuous, positive, nondecreasing function that would fail to converge? – Tartaglia's Stutter Mar 4 '19 at 12:25

Better write $$\sum^{\infty}_{k=1} f(k)$$ instead of $$\sum^{\infty}_{x=1} f(x)$$.
If $$b \in \mathbb N$$, then $$\sum^{b}_{k=1} f(k)$$ is a finite sum ! No convergence considerations are needed !
What do you mean by "$$\int^{b}_{a} f(x) dx \text{ converges}$$" ?
• But what if $f(x)$ has a vertical asymptote at at a point in the interval $[a,b]$. Can't the integral then go to infinity? – K. Claesson Mar 4 '19 at 12:40
• By saying $\int_{a}^{b} f(x) dx$ converges I mean that the integral has a finite value. – K. Claesson Mar 4 '19 at 12:47