Intro to Real Analysis: Show directly from the definition that the sequence is Cauchy.

Question

Using the definition we have to prove $(1+\frac{2}{2!}+......+\frac{2^n}{n!})$ is Cauchy.

So far I've gotten: If $m>n$ and $L=lim(x_n)$ then $|x_m-x_n|= \frac{2^{n+1}}{(n+1)!} +\cdots+\frac{2^m}{m!}$

It follows that: $|x_m-x_n|= \frac{2^{n+1}}{(n+1)!} +\frac{2^m}{m!} < \frac{2^{n+1}}{2^n} +\frac{2^m}{2^{m-1}}= \frac{1}{2^n}(2^{n+1} +\frac{2^m}{2^{m-n-1}})$

After this point, I can't figure out how to isolate for $\frac{2^n}{2^{n-1}}< $ $\epsilon$ or if I even took the right approach to solving this question.

Any help would be appreciated.

Answer 1

Hint: $$\frac{2^m}{m!} = \frac{2}{1}\,\frac{2}{2}\,\frac{2}{3}\cdots\frac{2}{m}$$ and almost all the factors of this product are $<1/2$ (and smaller).

Answer 2

$$|x_m-x_n|= \frac{2^{n+1}}{(n+1)!} +\cdots+\frac{2^m}{m!}$$ $$=\frac{2^{n}}{(n)!}\left(\frac{2}{n+1} +\cdots+\frac{2^{m-n}}{(n+1)\dots(m)}\right)$$

$$\le\frac{2^{n}}{(n)!}\left(\frac{2}{n} +\cdots+\frac{2^{m-n}}{n^{(m-n)}}\right)$$

$$\le\frac{2^{n}}{(n)!}\sum_{k=0}^\infty\left(\frac{2}{n}\right)^k$$

$\frac{2^{n}}{(n)!}$ can be made arbitrarily small by taking $n$ large, and the sum (which is convergent for $n>2$) only gets smaller as n increases.