Given a sequence $\{x_n\}$: $$ x_n = \sum_{k=1}^n a_kq^k $$ and: $$ \begin{cases} |a_k| \le C\\ |q| < 1\\ k\in\Bbb N \end{cases} $$ Prove $\{x_n\}$ is a fundamental sequence.
By definition of a fundamental sequence we want to show: $|x_n - x_m| < \epsilon$. So: $$ \begin{align} |x_n - x_m| &= \left|\sum_{k=1}^n a_kq^k - \sum_{k=1}^m a_kq^k\right| \stackrel{m>n}{=} \\ &= \left|\sum_{k=n+1}^m a_kq^k \right| \le \\ &\le \sum_{k=n+1}^m \left|a_kq^k \right| \le \\ &\le \sum_{k=n+1}^m \left|Cq^k \right| \stackrel{\text{geom. sum}}{=} \frac{Cq(q^m - q^n)}{q-1} \end{align} $$
Let $m > n$: $$ \frac{Cq(q^m - q^n)}{q-1} = \frac{Cq(q^n - q^m)}{1-q} \le \frac{Cq^{n+1}}{1-q} $$ Let $q$: $$ q = \frac{1}{1+r},\ r \in \Bbb R_{>0} $$ Then: $$ \frac{Cq^{n+1}}{1-q} = \frac{Cq}{(1-q)(1+r)^{n}} \stackrel{n > N}{\le} \frac{Cq}{(1-q)(1+r)^{N}} < \epsilon $$ Finally: $$ \frac{Cq}{(1-q)(1+r)^{N}} < \epsilon \iff (1+r)^{N} > \frac{Cq}{(1-q)\epsilon} \\ \iff N > \log_{1+r} \frac{Cq}{(1-q)\epsilon} \implies |x_n - x_m| < \epsilon $$
I'm kindly asking to verify the proof above. Thank you.
Well done.
If I want to nitpick, a very minor issue is that it is possible that $q$ can be negative. Hence, making this statement false.$$\sum_{k=n+1}^m \left|Cq^k \right| \stackrel{\text{geom. sum}}{=} \frac{Cq(q^m - q^n)}{q-1}$$
A quick fix is just to change $q$ to $|q|$.