Prove that $\sigma_n - e$ is decreasing faster than $e-S_n$ given specific $\sigma_n$ and $S_n$. 
Let $\sigma_n$ and $S_n$ be defined as:
  $$
\sigma_n = 3 - \sum_{k=1}^n\frac{1}{k(k+1)(k+1)!} \\
S_n = 1 + \sum_{k=1}^n\frac{1}{k!}
$$
  Show that $\sigma_n - e$ is decreasing faster than $e-S_n$.

I may use anything before the definition of a derivative.
Based on the question we want to eventually show that:
$$
\sigma_n - e \le e - S_n
$$
Lets adjust $\sigma_n$. Define $a_n$:
$$
a_n = \sum_{k=1}^n\frac{1}{k(k+1)(k+1)!}
$$
This may be expanded by partial fractions:
$$
\begin{align}
a_n &= \sum_{k=1}^n\frac{1}{k(k+1)(k+1)!} \\ 
&= \sum_{k=1}^n\frac{1}{k(k+1)}\cdot\frac{1}{(k+1)!} \\
&= \sum_{k=1}^n\left(\frac{1}{k} - \frac{1}{k+1}\right)\cdot\frac{1}{(k+1)!} \\
&= \sum_{k=1}^n\left(\frac{1}{k} - \frac{1}{k+1} - \frac{1}{(k+1)^2}\right)\cdot\frac{1}{k!} \\
&= \sum_{k=1}^n \frac{1}{kk!} - \color{red}{\sum_{k=1}^n \frac{1}{(k+1)k!}} - \color{blue}{\sum_{k=1}^n \frac{1}{(k+1)^2k!}} \\
&= \sum_{k=1}^n \frac{1}{kk!} - \color{blue}{\sum_{k=1}^n \frac{1}{(k+1)(k+1)!}} - \color{red}{\sum_{k=1}^n \frac{1}{(k+1)k!}} \\
\text{(telescoping)} &= 1 - \frac{1}{(n+1)(n+1)!} - \sum_{k=1}^n \frac{1}{(k+1)!}
\end{align}
$$
Now going back to the inequality:
$$
\sigma_n - e \le e - S_n \iff \\ 
\sigma_n + S_n \le 2e \iff \\
3 - a_n + S_n \le 2e
$$
Replacing the terms with actual sums one may obtain:
$$
\begin{align*}
3 - a_n + S_n &= 3 - \left(1 - \frac{1}{(n+1)(n+1)!} - \sum_{k=1}^n \frac{1}{(k+1)!}\right) + 1 + \sum_{k=1}^n{1\over k!} \tag{1.1} \\
&= 3 + \frac{1}{(n+1)(n+1)!} + \sum_{k=1}^n \frac{1}{(k+1)!} + \sum_{k=1}^n{1\over k!} \tag{1.2} \\
&= 3 + \frac{1}{(n+1)(n+1)!} + \sum_{k=2}^{n+1} \frac{1}{k!} + \sum_{k=1}^n{1\over k!} \tag{1.3} \\
&= 3 + \frac{1}{(n+1)(n+1)!} + \sum_{k=1}^{n} \frac{1}{k!} - 1 + {1\over (n+1)!} + \sum_{k=1}^n{1\over k!} \tag{1.4} \\
&= 2 + \frac{1}{(n+1)(n+1)!} + \frac{1}{(n+1)!} + 2\sum_{k=1}^{n} \frac{1}{k!} \tag{1.5} \\
&= 2\left(1 + \frac{1}{2(n+1)!}\left({1\over n+1} + 1\right) + \sum_{k=1}^{n} \frac{1}{k!}\right) \tag{1.6} \\
&= 2\left(\frac{(n+2)}{2(n+1)(n+1)!} + \sum_{k=0}^{n} \frac{1}{k!}\right) \tag{1.7}
\end{align*}
$$
Which eventually results into:
$$
\frac{(n+2)}{2(n+1)(n+1)!} + \sum_{k=0}^{n} \frac{1}{k!} \le e
$$
Now based on this question:
$$
e - S_n \le \frac{n+2}{(n+1)(n+1)!}
$$
While I wanted to show:
$$
\frac{n+2}{\color{red}{2}(n+1)(n+1)!} + S_n \le e \iff \\
\frac{n+2}{\color{red}{2}(n+1)(n+1)!} \le e - S_n
$$
Or summarizing:
$$
\frac{n+2}{2(n+1)(n+1)!} \le e - S_n \le \frac{n+2}{(n+1)(n+1)!}
$$
Which seems to be the case. Now I got stuck. How do I proceed from here?
Description of the steps:


*

*$(1.1)$ - replace $\sigma_n$ and $S_n$ with sums

*$(1.2)$ - cancel $-1+1$

*$(1.3)$ - change indexing in the sum

*$(1.4)$ - add and subtract $1$ and change the index of the sum. Fetch last term of the sum.

*$(1.5)$ - add the sums. 

*$(1.6)$ - factor out $2$. Factor out $1\over (n+1)!$ 

*$(1.7)$ - inject $1$ into the sum and change indexing. Cast brackets to a single fraction


Presuming my cursory review of your work leading up to the inequality $\frac{(n+2)}{2(n+1)(n+1)!} + \sum_{k=0}^{n} \frac{1}{k!} \le e$ didn't miss an error and assuming $n \in \mathbb N$, we may proceed as follows:
\begin{aligned}\\
e-\sum_{k=0}^{n} \frac{1}{k!}
&= \sum_{k=n+1}^{\infty} {1\over k!}\\
&\geq \frac1{(n+1)!}+\frac1{(n+2)!}\\
&\geq \frac1{2(n+1)!}+\frac1{(2n+2)(n+1)!} && (n+2\leq2n+2) \\
&=\frac{n+1}{2(n+1)(n+1)!}+\frac1{2(n+1)(n+1)!} \\
&=\frac{n+2}{2(n+1)(n+1)!}
\end{aligned}
Seems to me that you did most of the work in your original question.
