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My calculus textbook (as well as wikipedia and other online sources) list three conditions to verify before one can establish: $\sum\limits_{n=a}^\infty f(n)$ converges $\iff \int\limits_{a}^\infty f(x)dx$ converges:

1) $f(x)$ ($\mathbb{R}\rightarrow \mathbb{R}$) must be continuous on $[a,\infty)$

2) $f(x)$ must be non-increasing for $x$ sufficiently large.

3) $f(x)$ must be non-negative for $x$ sufficiently large.

Couldn't one instead just verify either:

$f(x)$ is continuous and eventually monotone OR $f(x)$ is monotone to arrive at the same conclusion?

If $f(x)$ is continuous and eventually monotone then there exists an integer $b>a$ such that $f(x)$ is monotone and doesn't change sign on $[b,\infty)$. If $f(x)$ is positive and non-decreasing on $[b,\infty)$ then both clearly diverge. If not, then either $f(x)$ or $-f(x)$ will satisfy the conditions for the Integral Test and we conclude. Continuity is merely required to ensure that $f(x)$ has finite Riemann integral on $[a,b]$.

If $f(x)$ is monotone, its Riemann integral exists over any compact interval so the indefinite integral can be expressed as a limit in the usual way. Choose an integer $b>a$ such that $f(x)$ doesn't change sign on $[b,\infty)$. Now proceed as before (noting that the proof for Integral Test doesn't technically require continuity on the non-increasing region).

I can't see where my argument goes wrong but if it's right, why are the conditions listed for the test redundant?

Terence C
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2 Answers2

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If $f$ is non-increasing and nonnegative then it is Riemann integrable and we have

$$\sum_{k=1}^{n-1} f(k) \geqslant\int_{1}^nf(x) \, dx \geqslant \sum_{k=2}^n f(k)$$

Note that $f(k)$ can always be defined since a monotone function has right and left limits at every point. This is enough to prove the integral test and it does not require $f$ to be continuous. This is easily generalized if $f$ is eventually non-increasing.

There are different /weaker conditions such that the integral test holds, where we do not even require $f$ to be monotone. See here.

I see redundant conditions for theorems often in elementary textbooks and, especially, online documents that may be subject to more relaxed standards. I can only speculate that it is an easy way to state a theorem correctly without delving too deeply in the subtleties of different combinations of weaker conditions.

RRL
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I am just now proving the following:

Given $n_0 \in \mathbb{N}$, $b \in \mathbb{R}$, and $f$ a eventually monotone real function, continuous on $[b,+\infty)$, if $\{a_n\}_{n = n_0}^{\infty} \in \mathsf{Seq}(\mathbb{R})$ is a restriction of $f$ (i.e., $a_n = f(n)$ for every natural $n$ sufficiently large), then the series $\sum_{n = n_0}^{\infty} a_n$ converges if and only if the improper integral $\int_{b}^{+\infty} f(x)\,dx$ converges.

I got here because I was wondering the same question as you: has anyone proved this more general version before? So, I went on Google and found your question.

My proof is too big, because I am writing it in great detail, and it is in portuguese, so I won't write it here. But it seems that you already have the idea to develop your own proof of this fact. I think the statement with the other condition ($f$ monotone, without continuity assumptions) is also true, as you said.

João Alves Jr.
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