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Let $S_n$ be a sequence defined by: $$ S_n = 1 + \sum_{k=1}^n {1\over k!} $$ Prove that: $$ e - S_n \le \frac{n+2}{n!(n+1)^2} $$

This problem comes in the limits section so i may use anything before the definition of a derivative. I've started with the following. Define: $$ L_n = \left(1+{1\over n}\right)^n $$

Rewrite $L_n$ as follows: $$ \begin{align} L_n &= \left(1+{1\over n}\right)^n \\ &= 1+ \sum_{k=1}^n {n\choose k}{1\over n^k} \\ &= 1 + \sum_{k=1}^n\frac{\overbrace{n\cdot(n-1)\cdot(n-2)\cdots(n-(k-1))}^{n - k + 1 \text{ times}}}{k!n^k} \end{align} $$

I've then tried to consider the difference between $L_n$ and $S_n$: $$ L_n - S_n = {1\over 2!}\left(1-{1\over n}\right) - {1\over 2!} + {1\over 3!}\left(1-{1\over n}\right)\left(1-{2\over n}\right) - {1\over 3!} + \cdots \\ = {1\over 2!}\left(1-{1\over n} - 1\right) + {1\over 3!}\left(\left(1-{1\over n}\right)\left(1-{2\over n}\right) - 1\right) + \cdots $$

Consider each parentheses. I've tried defining a sequence such that: $$ \forall n \ge 2:a_n = {1\over n!}\left(\prod_{k=1}^{n-1}\left(1-{k\over n}\right) - 1\right) $$

This sequence seems to always be less than $0$. So we have that: $$ \forall n\in \Bbb N :a_n \le 0 $$

Since $a_n$ is involved in $L_n - S_n$ we may also conclude that it is also less than $0$, namely: $$ L_n - S_n = \sum_{k=2}^na_n \le 0 $$

But at the same time: $$ \frac{n+2}{n!(n+1)^2} \ge 0 $$

So by this we have that: $$ L_n - S_n \le e - S_n \le 0 \le \frac{n+2}{n!(n+1)^2} $$

Now I don't see how to proceed. Should I bound the difference from another side? It feels like all i've done so far (above) is not even a usable argument for this problem, so could you please help me prove what's in the problem statement or point to the right direction.

I've made a Visualization of what is written above, perhaps that would be helpful.

roman
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2 Answers2

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We will use the following limit. If $S_n=\sum\limits_{k=0}^{n}\frac{1}{k!}$ then $$ \lim\limits_{n\rightarrow\infty} S_n = e. $$ This limit can be derived from $\lim\limits_{n\rightarrow\infty} \left(1+\frac{1}{n}\right)^n = e$.

Now we will prove that for any $m\geq 1$ inequality $S_{n+m}\leq S_n+\frac{n+2}{(n+1)(n+1)!}$ holds. Indeed, $$ S_{n+m}-S_n=\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\ldots+\frac{1}{(n+m)!}\leq \frac{1}{(n+1)!}\left(1+\frac{1}{n+2}+\ldots+\frac{1}{(n+2)^{m-1}}\right)=\frac{1}{(n+1)!}\cdot \frac{1-\frac{1}{(n+2)^m}}{1-\frac{1}{n+2}}<\frac{n+2}{(n+1)(n+1)!}. $$

From inequality $S_{n+m}\leq S_n+\frac{n+2}{(n+1)(n+1)!}$ when $m$ tend to infinity we get $e\leq S_n+\frac{n+2}{(n+1)(n+1)!}$ as desired.

richrow
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  • Thank you for the answer. Could you please justify the last step, it's not very obvious for me? namely $\frac{1}{(n+1)!} \cdot \frac{1- (n+2)^{-m}}{1-(n+2)^{-1}} < \frac{n+2}{(n+1)(n+1)!}$. – roman Jan 22 '19 at 14:08
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    Just note that $\frac{1}{(n+1)!} \cdot \frac{1- (n+2)^{-m}}{1-(n+2)^{-1}} <\frac{1}{(n+1)!} \cdot \frac{1}{1-(n+2)^{-1}}=\frac{n+2}{(n+1)(n+1)!}$ – richrow Jan 23 '19 at 18:29
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$\begin{array}\\ e-S_n &= \sum_{k=n+1}^{\infty} {1\over k!}\\ &= \frac1{(n+1)!}\sum_{k=n+1}^{\infty} {(n+1)!\over k!}\\ &= \frac1{(n+1)!}\sum_{k=0}^{\infty} {(n+1)!\over (n+1+k)!}\\ &= \frac1{(n+1)!}\left(1+\sum_{k=1}^{\infty} {(n+1)!\over (n+1+k)!}\right)\\ &= \frac1{(n+1)!}\left(1+\sum_{k=1}^{\infty} {1\over \prod_{j=1}^{k}(n+1+j)}\right)\\ &< \frac1{(n+1)!}\left(1+\sum_{k=1}^{\infty} {1\over \prod_{j=1}^{k}(n+2)}\right)\\ &= \frac1{(n+1)!}\left(1+\sum_{k=1}^{\infty} {1\over (n+2)^k}\right)\\ &= \frac1{(n+1)!}\left(1+\frac{1/(n+2)}{1-1/(n+2)}\right)\\ &= \frac1{(n+1)!}\left(1+\frac{1}{n+1}\right)\\ &= \frac1{(n+1)!}\left(\frac{n+2}{n+1}\right)\\ &= \frac{n+2}{(n+1)(n+1)!}\\ &= \frac{n+2}{(n+1)^2n!}\\ \end{array} $

marty cohen
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