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?

  • $\begingroup$ Can you come up with a finite sum, or a finite integral over a continuous, positive, nondecreasing function that would fail to converge? $\endgroup$ – 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 "$\int^{b}_{a} f(x) dx \text{ converges}$" ?

  • $\begingroup$ 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? $\endgroup$ – K. Claesson Mar 4 '19 at 12:40
  • $\begingroup$ Nevermind, if the function has a horizontal asymptote it will be discontinous at the point there the vertical asymptote exists. In that case, there probably are no integrals over a continous, positive, nondecreasing, finite interval that fail to converge. $\endgroup$ – K. Claesson Mar 4 '19 at 12:44
  • $\begingroup$ By saying $\int_{a}^{b} f(x) dx$ converges I mean that the integral has a finite value. $\endgroup$ – K. Claesson Mar 4 '19 at 12:47

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