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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!

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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 $n\to\infty$.

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  • $\begingroup$ Oh, I completely forgot of the ration test. Thank you! $\endgroup$ – roman Dec 26 '18 at 13:01
  • $\begingroup$ You're welcome ;) $\endgroup$ – Ankit Kumar Dec 26 '18 at 13:02

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