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
- $P_n$ is monotonic
- $P_n$ is bounded by $b$
- 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.
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 $$ \lim_{n\to\infty}P_n=b. $$