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Consider the series expansion for the exponential function at x = 1:

$$a_n := \sum\limits_{i = 0}^n{\frac{1}{i!}}$$

I want to prove that this is a Cauchy sequence, using the remainder formula for the exponential function. (That's the term for it I'm familiar with, anyway.) It would be nice if you could comment on whether this proof is correct or if it could be simplified without using additional theorems or such.

For the Cauchy criteria, for each $\epsilon > 0$ there must be a $N(\epsilon) \in N$ so that for all $m, n > N(\epsilon)$, $ |a_m - a_n| < \epsilon$.

So, take any $\epsilon > 0$. Now, choose $N(\epsilon)$ so that:

$${\frac{1}{(N+1)!}} < \frac{\epsilon}{2}.$$

(This is always possible because ${\frac{1}{n!}}$ goes to 0 when $n$ goes to infinity.)

Now, according to the partial sum formula, we got that:

$$\sum\limits_{i = 0}^{\infty}{\frac{1}{i!}} - \sum\limits_{i = 0}^{N(\epsilon)}{\frac{1}{i!}} = \sum\limits_{i = N(\epsilon)}^{\infty}{\frac{1}{i!}} < \epsilon$$

So if we choose any $m, n > N(\epsilon)$, and assume that $m > n$ (because when $m = n$, the Cauchy criteria is obviously true), it follows that:

$$ \sum\limits_{i = 0}^{m}{\frac{1}{i!}} - \sum\limits_{i = 0}^{n}{\frac{1}{i!}} = \sum\limits_{i = n}^{m}{\frac{1}{i!}} < \sum\limits_{i = N(\epsilon)}^{\infty}{\frac{1}{i!}} < \epsilon$$

which proofs the statement.

Edit: the remainder formula I'm familiar with and that I've been using states that:

$$\sum\limits_{i = 0}^{\infty}{\frac{1}{i!}} - \sum\limits_{i = 0}^{n}{\frac{1}{i!}} < 2 \frac{1}{(n+1)!} $$

if I recall correctly. Using this, can't we conclude that $\sum\limits_{i = N(\epsilon)}^{\infty}{\frac{1}{i!}} < \epsilon $ ?

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The easiest way I know to prove this is to use the fact that $$\forall n \in \Bbb N, \;\; 2^{n - 1} \ge n! $$

Now for any pair of natural numbers, $m \gt n $,

$$ |a_m - a_n| = \frac{1}{(n + 1)!} + ... + \frac{1}{(n + k)!} \;\;\;\;\; (k = m - n)$$

$$ \le \frac{1}{2^n} + ... + \frac{1}{2^{n + k - 1} } = \left({2 - \frac{1}{2^{n + k - 1}}}\right) - \left({2 - \frac{1}{2^{n - 1}}}\right) = \frac{1}{2^{n - 1}} \left({ 1 - \frac{1}{2^k}}\right) \lt \frac{1}{2^{n - 1}} \;\;\; \left({\text{using partial sums formula for geometric progressions: } \;\; S_n = \frac{a(1 - r^n)}{(1 - r)}}\right) $$

Now choose $ N(\epsilon) $ such that $$ 2^{N - 1} \gt \frac{1}{\epsilon} $$

$\mathscr{Q.E.D.}$

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