Let
$$S_{n} = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} + \cdots + (-1)^{n} \frac{1}{n!} = \sum_{k=2}^{n} (-1)^{k}\frac{1}{k!}.$$
a) What does $S_{n}$ converge to? Denote the limit point by $L$.
b) How quickly does it converge to $L$? For example, say something like, $|S_{n} - L| \sim \frac{1}{n}$.
For $(a)$, I'm pretty sure the answer is $e^{-1}$. I've got the answer to $(a)$ down, but I don't really know how to do part $(b)$.
By just plugging in values for $S_{n}$, it seems to converge pretty quickly, but I'm not able to provide a good estimation. Also, even if I am able to provide an estimation, I have no idea how to check whether or not it's correct. I was wondering if someone can please help me and show me how to check the rate of convergence (e.g. check if the guess is correct)
Thanks so much
You can get the convergence from the Taylor Series of $e^x,$ so the limit is correct.
I can't provide you with the least upper bound for the error, but I can tell you how to find a bound.
Implicitly $n\geq2$. We compute $$\left|S_n-L\right|=\left|\sum_{k=n+1}^\infty(-1)\frac{1}{k!}\right|=\frac{1}{n!}\left|\frac{1}{n+1}-\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}-\ldots\right|$$ $$\leq\frac{1}{n!}\left(\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\ldots\right)$$ $$\leq\frac{2}{n!(n+1)}\left(\frac{1}{2}+\frac{1}{2(n+2)}+\frac{1}{2(n+2)(n+3)}+\ldots\right)$$ $$|S_n-L|\leq\frac{2(e-1)}{n!(n+1)}=\frac{2(e-1)}{(n+1)!}.$$