Show that the interest rate sequence is Cauchy

Question

I have a problem where I have to show that the interest rate sequence $\{x_n\}$ defined by

$x_n := (1+\frac{1}{n})^n$

is Cauchy without using that $\{x_n\}$ is convergent.

I think I can apply the binomial theorem to show that but I have no idea how to do that.

Thank you

Answer 1

Take $m = n+1$
So, $$|x_m-x_n| = |x_{n+1}-x_n| = |(1+\frac{1}{n+1})^{n+1}-(1+\frac{1}{n})^n|$$ $$=|[1+(n+1)\cdot \frac{1}{n+1}+\binom{n+1}{2}(\frac{1}{n+1})^2+\binom{n+1}{3}(\frac{1}{n+1})^3+...+\binom{n+1}{n+1}(\frac{1}{n+1})^{n+1}]-[1+(n)\cdot \frac{1}{n}+\binom{n}{2}(\frac{1}{n})^2+\binom{n}{3}(\frac{1}{n})^3+...+\binom{n}{n}(\frac{1}{n})^{n}]|$$ $$=|[1 + 1 + \frac{1}{2!}(1-\frac{1}{n+1}) + \frac{1}{3!}(1-\frac{1}{n+1})(1-\frac{2}{n+1}) +...+\frac{1}{(n+1)!}(1-\frac{1}{n+1})(1-\frac{2}{n+1})(1-\frac{3}{n+1})...(1-\frac{n+1}{n+1})]-[1+1+\frac{1}{2!}(1-\frac{1}{n}) + \frac{1}{3!}(1-\frac{1}{n})(1-\frac{2}{n}) +...+\frac{1}{(n)!}(1-\frac{1}{n})(1-\frac{2}{n})(1-\frac{3}{n})...(1-\frac{n}{n})]|$$ $$=|\frac{1}{2!}(\frac{1}{n}-\frac{1}{n+1})+\frac{1}{3!}[(1-\frac{1}{n})(1-\frac{2}{n})-(1-\frac{1}{n+1})(1-\frac{2}{n+1})]+...|$$ Now we know, $n!>2^n \text{for n} \ge 2 \,\ \text{and} \,\ \frac{1}{n!} < \frac{1}{2^n}$ $$<|\frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + ... + \frac{1}{2^n} + \frac{1}{2^{n+1}}|$$ $$=|\frac{1}{4}\left( \frac{1-(\frac{1}{2})^{n}}{1-\frac{1}{2}}\right)|$$
$$=|\frac{1}{2}-(\frac{1}{2})^n|$$ $$<|\frac{1}{2}|$$ $$<\epsilon$$ for $\epsilon > \frac{1}{2}$

Hope this helps in proving sequence is cauchy.