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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.

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

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

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

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

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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).

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    $\begingroup$ was about to write a more powerful hint of the same kind. $\endgroup$ – Exodd Mar 1 '18 at 22:53
  • $\begingroup$ @Exodd, write it. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 1 '18 at 22:56

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