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Let $$ \Delta P = (b-P_n)/2 \\ P_{n+1} = P_n + \Delta P\\ P_n : \mathbb{Q} \to \mathbb{R} $$ We shall attempt to prove that $$ \lim_{n \to \infty} P_n = b $$ I'll try in three parts

  1. $P_n$ is monotonic
  2. $P_n$ is bounded by $b$
  3. Something along the lines of $\sup P_n = b$

First, prove $P_n$ is bounded by $b$. Assume the contrary by taking $P_n$ to be the supremum of $P_n < b$ and therefore $P_{n+1} > b$. Now that implies $\Delta P > b-P_n$ which violates the definition of the recurrence. Therefore, $P_n$ is bounded from above by $b$

Second, let us prove that $P_n$ is monotonic with respect to $n$. Assuming $b > P_0$, $P_1 > P_0$. Now choose an arbitrary $P_{n}$, where $b > P_n$ from the previous result. If $P_{n+1} = P_n + \Delta P$, $\Delta P = \frac{b-P_n}{2} > 0$ and therefore $P_n$ is increasing for $P_n < b$

Now let us prove that $\lim_{n \to \infty} P_n = b$.

That means that for all $M, \epsilon > 0$ $$ n > M \implies |P_n - L| < \epsilon $$ To start, we know $L \le b$ by our first result.

Now all we must prove is that $L < b$ is impossible, which is what I am having trouble with.

What I am having trouble with

Well if for all $n>M$, $P_n < L$ then for all $P_n$, $$ \Delta P = \frac{b-P_n}{2}< L-P_n < \epsilon $$ I'm having trouble constructing a contradiction and was hoping for some help doing so.

I specifically want to use a delta epsilon proof and not any other form of proof.

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    Well yes. it is the same question. But I asked it again in a different form this time requiring a delta epsilon proof and not just an arbitrary proof. The goal for my choosing to do this exercise is not to see how to prove this recurrence converges. It is to understand delta epsilon proofs more formally. – 2017-02-17

1 Answers 1

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Note $$ P_{n+1}=P_n+\Delta=P_n+\frac{b-P_n}{2} $$ and hence $$ P_{n+1}-b=\frac{1}{2}(P_n-b). $$ Thus $$ P_n-b=(\frac12)^{n-1}(P_1-b) $$ which implies $$ \lim_{n\to\infty}P_n=b. $$