The problem I am trying to take is:
Prove that $x_n=\frac{a_1}{e^1}+\frac{a_2}{e^2}+...+\frac{a_n}{e^n}$ is a Cauchy sequence s.t. $|a_n|\leq1000$
I begin this like other Cauchy sequences proofs, let $n,m$ be some integers and I want to find a minimum for a particular $n_0$ s.t. $\forall m,n\geq n_0 \implies |x_n-x_m|<\epsilon$ where $\epsilon>0$. Here is my attempt: (assuming $n<m$ although this should not affect the basis of the proof):
\begin{align} |x_n-x_m|\\ =\Biggl|\frac{a_1}{e^1}+\frac{a_2}{e^2}+...+\frac{a_n}{e^n}-\Biggl(\frac{a_1}{e^1}+...+\frac{a_n}{e^n}+...+\frac{a_m}{e^m}\Biggl)\Biggl|\\ =\Biggl|\frac{a_{n+1}}{e^{n+1}}+...+\frac{a_m}{e^m}\Biggl|\\ \leq\Biggl|\frac{1000}{e^{n+1}}+...+\frac{1000}{e^m}\Biggl|<\epsilon \end{align}
At this point I get a little stuck on how to get a nice lower bound of $n_0$. Does anyone have any hints on how to proceed?
We have $$ |x_n - x_m| \leq \sum_{k=n+1}^m \frac{1000}{e^k} = \frac{1000}{e^n} \sum_{k=1}^{m-n} \frac{1}{e^k} $$ but $$ \sum_{k=1}^{m - n} \frac{1}{e^k} < \sum_{k=1}^{m-n} \frac{1}{2^k} < 1 $$ so $$ |x_n - x_m| \leq \frac{1000}{e^n} $$ so the difference can be arbitrarily small.