let $\{x_n\}$ be a sequence of real numbers s.t $|x_{n+1} - x_n| \leq \alpha ^n$ $\forall n \in \mathbb{N}$ where $0 < \alpha < 1$
a) prove $|x_m - x_n | \leq \dfrac{\alpha^n}{1-\alpha} $ for $m>n$
done part a
b) Prove $|x_m - x_n | \leq \dfrac{\alpha^m + \alpha^n}{1-\alpha}$ for all $m,n$
For part b), is this the correct way to go about it?
for $m>n$, $|x_m - x_n | \leq \dfrac{\alpha^n}{1-\alpha} $ for $n>m$ $|x_m - x_n | = |x_n - x_m | \leq \dfrac{\alpha^m}{1-\alpha}$
so adding these $\forall m,n$ $|x_m - x_n| \leq \dfrac{\alpha^m + \alpha^n}{2(1-\alpha)} \leq \dfrac{\alpha^m + \alpha^n}{1-\alpha}$? Is this the correct way to do it?
c) Prove that $\{x_n\}$ is a cauchy sequence.
given epsilon, choose $N$ s.t. $\dfrac{\alpha^N + \alpha^N}{1-\alpha} < \epsilon$ hence for $m,n>N$ by the previous parts of the questions $|x_m - x_n| \leq \dfrac{\alpha^m + \alpha^n}{1-\alpha} \leq \dfrac{\alpha^N + \alpha^N}{1-\alpha} < \epsilon$ so the sequence is cauchy.
Is this correct?
Part b: you discussed two unique different cases: for all m,n, $$|x_m - x_n| \leq \color{red} {\dfrac{\alpha^m + \alpha^n}{2(1-\alpha)}} \leq \dfrac{\alpha^m + \alpha^n}{1-\alpha}? $$ so it is correct except red part(delete it).
Part c: $$\lim _{n \to \infty}|x_{n+1} - x_n| \leq \lim _{n \to \infty}\alpha ^n=0 $$ since $0 < \alpha < 1$ the result.
Uh one more note for Part B $$|x_m - x_n | \leq \dfrac{ \alpha^n}{1-\alpha}\leq \dfrac{ \alpha^n}{1-\alpha}+\color{red}{\underbrace{\dfrac{ \alpha^m}{1-\alpha}}_{\text{positive term}}}$$ again the result