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

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.

• I think you should correct first the difference – Exodd Mar 1 '18 at 22:46
• Your bounding is too coarse: $\frac{2^n}{2^{n-1}} = 2\not<\epsilon$ for $\epsilon$ small. – Martín-Blas Pérez Pinilla Mar 1 '18 at 22:47
• also, your sum doesn't start from 1, but 2, or your second term is 2/1! – Exodd Mar 1 '18 at 22:54

$$|x_m-x_n|= \frac{2^{n+1}}{(n+1)!} +\cdots+\frac{2^m}{m!}$$ $$=\frac{2^{n}}{(n)!}\left(\frac{2}{n+1} +\cdots+\frac{2^{m-n}}{(n+1)\dots(m)}\right)$$
$$\le\frac{2^{n}}{(n)!}\left(\frac{2}{n} +\cdots+\frac{2^{m-n}}{n^{(m-n)}}\right)$$
$$\le\frac{2^{n}}{(n)!}\sum_{k=0}^\infty\left(\frac{2}{n}\right)^k$$
$\frac{2^{n}}{(n)!}$ can be made arbitrarily small by taking $n$ large, and the sum (which is convergent for $n>2$) only gets smaller as n increases.
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).