If $|p_{n+2} - p_{n+1}| \le \frac{1}{3}|p_{n+1}-p_n|$, then prove that $p_n$ is a Cauchy sequence. (Hint: Use geometric series)

Question

If $|p_{n+2} - p_{n+1}| \le \frac{1}{3}|p_{n+1}-p_n|$, then prove that $p_n$ is a Cauchy sequence. (Hint: Use geometric series)

For $n = 1$, $|p_3 - p_2| \le \frac{1}{3} |p_2 - p_1|$

For $n = 2$, $|p_4 - p_2| \le \frac{1}{3} |p_3 - p_2|$ then $|p_4 - p_2| \le \frac{1}{3} |p_3 - p_2|\le \frac{1}{3^2} |p_2 - p_1|$

So for n-term, it will be $|p_n - p_{n-1}| \le \frac{1}{3^n} |p_2 - p_1|\lt \frac{1}{3^n} |p_2| + \frac{1}{3^n}|p_1| $

I don't know how to get to $|p_n - p_m| \lt \epsilon$ from that but that's all I get.

Answer 1

You've proved that $$\lvert p_n - p_{n-1} \rvert \le \frac{C}{3^n} $$ where $C = \lvert p_2 - p_1 \rvert$. Then for $n > m$, by adding and subtracting all the terms $p_j, m < j < n$ and using the triangle inequality, we have \begin{align*} \lvert p_n - p_m \rvert &= \lvert p_n - p_{n-1} + p_{n-1} - \cdots - p_{m+1} - p_{m+1} - p_m \rvert \\ &\le \sum^n_{j=m+1} \lvert p_j - p_{j-1}\rvert \\ &\le \sum^n_{j=m+1} \frac{C}{3^{j}} \\ &\le \sum^\infty_{j=m+1} \frac{C}{3^{j}} = C \frac{\tfrac{1}{3^{m+1}}}{1 - \tfrac 1 3} = \frac 32 C \left( \frac 1 3 \right)^{m+1} \to 0 \,\,\,\,\,\,\,\, \text{ as } m \to \infty. \end{align*} Thus taking $n,m$ sufficiently large, we can ensure this is less than any given $\epsilon > 0$.

Answer 2

You are provided $$|p_{n+2} - p_{n+1}| \le \frac{1}{3}|p_{n+1}-p_n|\le\frac1{3^2}|p_n-p_{n-1}|\le\dots\le\frac1{3^n}|p_2-p_1|.$$

Now, suppose $|p_2-p_1|=k$.

Now, consider $m= \lfloor\log_3\left(\frac k\varepsilon\right)\rfloor$.

Then you have $$|p_{n+2} - p_{n+1}|\le \varepsilon\quad \forall n\ge m.$$