# Show that $S_n=1+{x\over1!}+{x^2\over2!}+\cdots+{x^n\over n!}$ converges for $n\in\Bbb N,\ x \in\Bbb R$ without using Taylor series.

Given a sequence $$\{S_n\}$$, $$n\in\Bbb N$$: $$S_n=1+{x\over1!}+{x^2\over2!}+\cdots+{x^n\over n!}$$ Prove that $$S_n$$ converges for all $$x\in\Bbb R$$.

Please note that i know $$S_n$$ is a simple Taylor series for $$e^x$$, the case is I'm not supposed to know (and/or use) that fact when solving this problem since derivatives have not been defined yet.

I've started with a simpler case assuming $$x = 1$$. So the sequence becomes: $$S_n = 1+{1\over1!}+{1\over2!}+\cdots+{1\over n!}$$

It seems reasonable to use Cauchy Criterion for the sequence. Namely suppose $$m>n$$, then we want to show: $$|x_n - x_m| < \epsilon \\ |x_m - x_n| = \left|\sum_{k=n+1}^m {1\over k!}\right|$$ Lets try to synthetically bound the sum by: \begin{align} {1\over k!} &= {1\over k!}\left(1 - {1\over k} + {1\over k}\right) \\ &\le {1\over k!}\left(1 + {1\over k-1} - {1\over k}\right) \\ &= {1\over k!}\left({k\over k -1} - {1\over k}\right) \\ &= {1\over (k - 1)(k-1)!} - {1\over k\cdot k!} \end{align}

By telescoping we obtain: $$\left|\sum_{k=n+1}^m {1\over k!}\right| \le \left|{1\over n\cdot n!} - {1\over m\cdot m!}\right|$$

Given $$m>n$$ and $$n,m \in \Bbb N$$: $$\left|{1\over n\cdot n!} - {1\over m\cdot m!}\right| \le \left|1\over n\cdot n!\right| = {1 \over n\cdot n!}$$

Applying the limit to $$|x_m - x_n|$$ one may obtain: $$0 \le \lim_{n\to\infty}|x_{n+p} - x_n| \le \lim_{n\to\infty} {1\over n\cdot n!} = 0$$

So squeezing $$|x_m - x_n|$$ gives: $$\lim_{n\to\infty}|x_{n+p} - x_n| = 0,\ p\in \Bbb N$$

Which would eventually mean: $$|x_m - x_n| < \epsilon$$

However I'm not sure how to find the index $$N_\epsilon$$ from which the inequality becomes true since the expression for the upper bound involves a factorial.

If we now put $$x = x_0 \in \Bbb R$$: $$0 \le \lim_{n\to\infty}|x_{n+p} - x_n| \le \lim_{n\to\infty} {x_0\over n\cdot n!} = 0$$ Which doesn't influence the value of the limit.

There are three questions in my mind:

1. Is the overall reasoning valid?
2. Is it possible to find a closed form of $$N(\epsilon)$$, such that $$n, m > N_\epsilon \implies |x_n - x_m| < \epsilon$$?
3. Should I consider two cases for $$x\ge 0$$ and $$x<0$$

Thank you!

You can simply use ratio test to show that the given series converges. Any sequence $$S$$ is converging if $$r=\lim_{n\to \infty}\frac{a_{n+1}}{a_n}<1$$ For this problem, $$r=\frac{x^{n+1}}{(n+1)!}\frac{n!}{x^n}$$ $$=\frac{x}{n+1}$$ $$r$$ goes to $$0$$ as $$n\to\infty$$.