I'm having trouble with another analysis homework problem:
Let $x_n$ be a sequence in $\mathbb{R}$ such that $d(x_n, x_{n+1}) \le \frac{d(x_{n-1},x_n)}{2}$. Show that $x_n$ is a Cauchy sequence.
I know that I need to prove that for any $\epsilon$ there exists a $N$ such that $m,n \ge N$ implies $d(x_m,x_n) < \epsilon$. However, I am having a hard time defining this $N$. Here is what I have so far:
$$d(x_n, x_{n+1}) \le \frac{d(x_{n-1},x_n)}{2} \le d(x_{n-1},x_n) \le d(x_{n-1},x_{\lambda}) + d(x_{\lambda}, x_n).$$
Now, if I can show that for any $\epsilon$, there exists a $N_1$ such that $m,n \ge N_1$ implies $d(x_m,x_n) < \frac{\epsilon}{2}$ and for any epsilon, there exists a $N_2$ such that $p,q \ge N_2$ implies $d(x_p,x_q) < \frac{\epsilon}{2}$, then I can simply define $N = \max\{N_1,N_2\}$ and I am done. Unfortunately, I do not know how to go about finding $N_1$ and $N_2$.
I'll give it a shot. Fix $\epsilon>0$. Let $N$ be such that $\frac{d(x_{0},x_{1})}{2^{N}}<\epsilon$. Then, for $n,m > N+1$, we have $m=n+k$ for some positive $k$ and:
$$ d(x_{n},x_{m})\leq d(x_{n},x_{n+1})+\ldots+d(x_{m-1},x_{m}) \leq \frac{d(x_{0},x_{1})}{2^{n}} + \frac{d(x_{0},x_{1})}{2^{n+1}} + \ldots + \frac{d(x_{0},x_{1})}{2^{n+k-1}} \\ \leq d(x_{0},x_{1})\frac{1}{2^{n}}(\frac{1-(1/2)^{k}}{1/2}) = d(x_{0},x_{1})\frac{1}{2^{n-1}}({1-(1/2)^{k}}) \leq \frac{d(x_{0},x_{1})}{2^{n-1}} < \epsilon $$
since $n >N+1$. I think this is it, but feel free to judge my answer.