How do you come to the conclusion $0<1/(m+1)!+\dots+1/n!<1/(2^n)$ from $2^k<k!$ if $k\geq 4$?

Question

I'm working out of the Bartle and Sherbert Introduction to Real Analysis book and I'm looking at the partial solution to proving the sequence $(1+1/2!+\dots+1/n!)$ is Cauchy (Exercise 3.5.2(b)). Their partial solution was:

$0<1/(n+1)!+\dots+1/n!<1/(2^m), $ since $2^k<k!$ for $k\geq 4$.

I'm not understanding how they came to this conclusion. It seems that (as they tend to do in Cauchy proofs) they are assuming $n,m\in\mathbb{N}$ such that $m>n.$ I'm pretty sure an exercise from a few chapters ago showed $2^k<k!$ for $k\geq 4$ by induction, so I understand why they said that. I don't know how assuming $m>n$ along with stating that $2^k<k!$ for $k\geq 4$ implies that $1/(n+1)!+\dots+1/m!<1/(2^n)$. I get how one of the terms in the inequality would be less than $1/(2^n)$ (e.g. $1/(n+1)!<1/2^n$), but I'm not sure how all the terms added up together would be less than $1/2^n.$ Could anyone shed light on how this would work? Thank you!

PS: I understand that the proof that $(1+1/2!+\dots+1/n!)$ is Cauchy follows: Suppose $\epsilon>0.$ Then by the Archimedean Property there exists some $H=H(\epsilon)\in\mathbb{N}$ such that $1/H<\epsilon.$ Then for all $m,n\in\mathbb{N}$ such that $m>n\geq H,$

$$|(1+1/2!+\dots+1/m!)-(1+1/2!+\dots+1/n!)|=1/(n+1)!+\dots+1/m!<1/(2^n)\leq 1/H<\epsilon. $$

Since $\epsilon$ is arbitrary, we conclude by definition of Cauchy sequences of real numbers that the sequence $(1+1/2!+\dots+1/n!)$ is Cauchy.

So the only part I don't understand is the quote above.

Answer 1

You're mixing up some of your $m$ and $n$, typing one when you mean the other. But that's a minor detail you can fix on your own.

The main idea is simply that, for sufficiently large $k$, $$\frac{1}{2^k} > \frac{1}{k!},$$ hence we may replace each factorial term with the respective power of $2$:
$$\begin{align} \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \cdots + \frac{1}{m!} &< \frac{1}{2^{n+1}} + \frac{1}{2^{n+2}} + \cdots + \frac{1}{2^m} \\ &< \frac{1}{2^{n+1}} + \frac{1}{2^{n+2}} + \cdots \\ &= \frac{1}{2^{n+1}} \left(\frac{1}{1 - 1/2}\right) \\ &= \frac{1}{2^n}. \end{align}$$ In the first step, we bounded each term by the respective power of $2$. In the second step, we took the resulting finite geometric series and bounded it from above by an infinite geometric series, which contains all of the terms in the finite series but also additional small terms $1/2^{m+1} + 1/2^{m+2} + \cdots$. Then we sum the infinite series and get $1/2^n$ as claimed.

If using an infinite geometric series does not sit well with you, one can just as easily explicitly compute the finite sum:

$$\sum_{k=n+1}^m \frac{1}{2^k} = \frac{2^m - 2^n}{2^{m+n}} = \frac{1}{2^n} - \frac{1}{2^m}$$ and again this is strictly smaller than $1/2^n$ for any $m > 0$.

Answer 2

I would do this:

$\begin{array}\\ \sum_{k=n+1}^m \dfrac1{k!} &=\dfrac1{n!}\sum_{k=n+1}^m \dfrac1{k!/n!}\\ &=\dfrac1{n!}\sum_{k=n+1}^m \dfrac1{\prod_{j=n+1}^k j}\\ &\lt\dfrac1{n!}\sum_{k=n+1}^m \dfrac1{\prod_{j=n+1}^k 2}\\ &=\dfrac1{n!}\sum_{k=n+1}^m \dfrac1{2^{k-n}}\\ &=\dfrac1{n!}\sum_{k=1}^{m-n} \dfrac1{2^{k}}\\ &\lt\dfrac1{n!}\sum_{k=1}^{\infty} \dfrac1{2^{k}}\\ &=\dfrac1{2n!}\\ \end{array} $

This will give you what you want.

You can easily do better, but this is enough.