0

The following is the proof of the Integral Test from Spivak's Calculus - 2nd edition - Ch.22.

enter image description here

I understand the majority of the proof, but I'm having trouble wrapping my head around the claim of:

"The existence of $\lim_{A \to \infty}\int_{1}^{A}f$ is equivalent to the convergence of the series $$\int_{1}^{2}f + \int_{2}^{3}f + \int_{3}^{4}f + \dots$$"

Since I was having trouble with the concept I've attempted to prove this claim, but it is here where I'm running into problems. So given the above conditions of the original theorem I could interpret $\int_{1}^{A}f$ as being the terms of a sequence. So let's define the terms as: $b_{A} = \int_{1}^{A}f$.

To continue on I asked myself "What does it mean for the series to converge?" Here it means that if I define $S_{A} = b_{1} + b_{2} + \dots + b_{A}$ (partial sum), then $$\lim_{A \to \infty}S_{A} = \sum_{A = 1}^{\infty}b_{A} = l$$ for some $l$.

From here the question I ask myself is "How to show a sequence converges?" from which I would answer with the standard definition of: For all $\epsilon > 0$ there is a natural number $A > 0$ such that for all $a \in \mathbb{N}$, if $a > A$, then $|S_{a} - l| < \epsilon$. But this probably won't help me in this situation. I looked up previous proofs on the integral test and one of them dealt with this issue:

Proof of integral test

but in that version the OP had the additional fact of: $\sum_{k=2}^n f(k) \le \int_1^n f(x) dx \lt \int_1^{\infty} f(x) dx \lt \infty$

I don't have this fact at my disposal. But I'm also attempting to prove this from my own formulation of the claim so perhaps I would've needed that to prove the original claim. But still I'm stuck on showing the original claim. What can I do to prove this?

EDIT: In the question I have incorrectly defined my $b_{n}$. The correct way they should be defined is in the comments, I'm leaving the mistake in the question in case others in the future make the same mistake.

D.C. the III
  • 5,619
  • 1
    You do have that fact - use the fact that $f$ is (1.)positive, (2.) decreasing, (3.) Each region of integration has length $n+1 - n = 1$ – rubikscube09 Oct 07 '20 at 20:40
  • How do you know each region of integration is of length $1$? – D.C. the III Oct 07 '20 at 20:44
  • 1
    @dc3rd Your sum of the $b_i$ is of the form $\int_1^2 f + \int_1^3 f + \cdots,$ when it should instead be of the form $\int_1^2 f + \int_2^3 f + \cdots.$ – Ben Grossmann Oct 07 '20 at 20:44
  • @BenGrossmann I see what you mean. Could I "possibly" rectify that by defining my sequence as $c_{A} = b_{A} - b_{A-1}$? as an example term: $c_{3} = b_{3} - b_{2} = \int_{1}^{3}f - \int_{1}^{2}f = \int_{1}^{3}f + \int_{2}^{1}f = \int_{2}^{3}f$ (flipping the integral). Or is this just over complicating things? – D.C. the III Oct 07 '20 at 20:53
  • @dc3rd Yes. Equivalently, you would be defining $c_A = \int_{A-1}^A$. – Ben Grossmann Oct 07 '20 at 21:02
  • @BenGrossmann. Cool....But I'm still stuck on how to get this claim. – D.C. the III Oct 07 '20 at 21:04

1 Answers1

2

Hint: Break up your proof of equivalence into a proof of "if" and "only if". You should find it easy to prove that if $\lim_{A \to \infty} \int_1^A f$ exists, then the sum $\int_1^2 f + \int_2^3 f + \cdots$ also converges.

For the other direction, you must use the fact that $f$ is positive and decreasing.

As for the "fact that OP had", you would only need to prove that $f(k) \leq \int_{k-1}^{k} f(x)\,dx$.

Ben Grossmann
  • 225,327
  • I'm going to give it a go and will report back. – D.C. the III Oct 07 '20 at 21:12
  • I think I figured out the first direction. I'll show the rough work because that would imply i I treated it right. I need to show $|S_{A} - l| < \epsilon$. Using how I defined $c_{A}$ in the comments, if I work backwards: $|S_{A} - l| = |\sum_{i = 1}^{A}c_{i} - l| = |c_{1} + c_{2} + \dots + c_{A} - l| = |b_{1} - b_{0} + b_{2} - b_{1} + \dots + b_{A} - b_{A-1} - l| = |b_{A} - b_{0} - l| < |b_{A} - l| < \epsilon$. The $|b_{A} - l| < \epsilon$ comes from us being given $\lim_{A \to \infty}\int_{1}^{A}f = \lim_{A \to \infty}b_{A}$ exists. So to do the proof I would start with the last part. – D.C. the III Oct 07 '20 at 21:58
  • As for the second direction, I'm still a little stuck. – D.C. the III Oct 07 '20 at 21:59