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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
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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.

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    $\begingroup$ Wonderful. Thank you! $\endgroup$ – roman Jan 24 at 23:10

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