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I'm taking Calculus II next semester so I don't have all the tools from it to back me up but something came up in a discussion this evening that asked me to prove that the following sequence converges. As I'm new to formal math I'm hoping someone can verify my proof, especially step $6$.

I must prove that

$$\lim_{n\to\infty}P_n=b$$

where

$$P_{n+1} = P_n + \frac{b-P_n}{2}$$

$$P_0 = a$$

and $a,b \in \mathbb{R}$ such that $a>b$

My proof goes as follows:

  1. We must prove that $$\lim_{n\to\infty}P_n=b$$
  2. That is, $\forall\epsilon>0$ there exists a $\delta$ such that there exists an $n$ satisfying $$n>\delta \iff b-P_n<\epsilon$$

  3. First, let us prove that $P_n$ is bounded by $b$ by contradiction. Let us choose an $i$ such that $P_i < b$ and $P_{i+1}>b$. Then we have $P_{i+1} = P_i + \Delta P$ where $\Delta P > b-P_i$. but this is impossible because by the definition of the recurrence $\Delta P = \frac{b-P_i}{2} < b-P_i$. Therefore, $P_i$ is bounded from above by $b$.

  4. Additionally, it is trivial to show that $P_{i+1} > P_i$.

  5. With this in mind, choose an arbitrary $\epsilon$. We need to show that there exists an $n$ such that $P_n > b-\epsilon$. Assume the contrary, that $P_n$ is bounded by $\epsilon$.

  6. Using a similar argument to what which was used before, assume $P_n$ is bounded by $\epsilon$ and then there exists a $P_i$ such that $P_{i+1}\not > P_i$. But this results in a contradiction with (4)
  7. Therefore, $P_n$ is not bounded by $\epsilon$ and we have proved what we want to prove.
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    Additionally, you can show the sequence is Cauchy. Then, along with monotone convergence, you will get convergence (then, you can just substitute $L$ for the limit in the recurrence relation). – 2017-02-16
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    Unless you *have* to use an $\epsilon/\delta$ argument, why not just write it as $P_{n+1} = \frac{1}{2}P_n+\frac{1}{2}b\,$, then telescope to find the general term for $P_n\,$, then its limit. – 2017-02-16
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    (1) Not familiar with Cauchy. Only have Calc I and 0.3 Calc III under my narrow undergraduate belt. (2) I'd prefer to use an $\epsilon/\delta$ argument. – 2017-02-16
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    For when it is impossible to solve the recurrence relation – because I want this proof to generalize to all monotonic relations bounded from above. – 2017-02-16
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    You're basically halving the distance between A and B – 2017-02-16

1 Answers 1

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The direct proof is to subtract $b$ from both sides, which gives $P_{n+1}-b=\frac{1}{2}(P_{n}-b)\,$. Iterating the latter gives $P_{n+1}-b=\frac{1}{2^{n+1}}(P_0-b)=\frac{a-b}{2^{n+1}}\,$. Use your favorite proof, $\epsilon/\delta$ or other, to show that $\frac{a-b}{2^{n}} \to 0\,$, then it follows that $P_{n}-b \to 0 \iff P_n \to b\,$, and that holds true regardless of $P_0$.

Following is a summary critique of the posted proof.

  1. That is, $\forall\epsilon>0$ there exists a $\delta$ such that there exists an $n$ satisfying $$n>\delta \iff b-P_n<\epsilon$$

That should be $|b-P_n|<\epsilon\,$. And it's an $\implies$ implication, not an $\iff$ equivalence. There could well be other terms $P_n$ with $n \le \delta$ such that $|b-P_n|<\epsilon\,$.

  1. With this in mind, choose an arbitrary $\epsilon$. We need to show that there exists an $n$ such that $P_n > b-\epsilon$.

This step is relying on $P_n$ having been already proved to be monotonic and bounded, but those assumptions could be better spelled out.

Assume the contrary, that $P_n$ is bounded by $\epsilon$.

It is hard to tell what "$\,P_n$ is bounded by $\epsilon\,$" means.

  1. Using a similar argument to what which was used before, assume $P_n$ is bounded by $\epsilon$ and then there exists a $P_i$ such that $P_{i+1}\not > P_i$. But this results in a contradiction with (4)

It is even harder to tell how that supposed contradiction is derived.