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A bit rusty on this stuff. The whole problem is proving this is true:

$ 2 \leq 1 + \sum_{m=1}^{n} \frac{1}{m!} \leq 1 + \sum_{m=1}^{n} \frac{1}{2^{m-1}} < 3. $

I have figured out the first two inequalities:

$2 \leq 1 + \sum_{m=1}^{n} \frac{1}{m!}, \quad \quad \text{and}$ $1 + \sum_{m=1}^{n} \frac{1}{m!} \leq 1 + \sum_{m=1}^{n} \frac{1}{2^{m-1}} .$

But I am having trouble proving the last bit:

$1 + \sum_{m=1}^{n} \frac{1}{2^{m-1}} < 3 .$

Any tips?

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    Informally, look at $1$, $1+\frac{1}{2}$, $1+\frac{1}{2}+\frac{1}{4}$, $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}$, and so on. At each step, our distance from $2$ is divided by $2$. If this is not clear, a picture helps.2011-11-14

2 Answers 2

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The key thing you need is the fact that $ \sum_{m=1}^n a^{m-1} = \sum_{m=0}^{n-1} a^m = \frac{a^n-1}{a-1}$ For you, $a = \frac{1}{2}$. So, $ \frac{(1/2)^n -1}{(1/2)-1} = 2(1-(1/2)^n)$ Notice that that quantity is just a little less than 2. So 1 plus the sum is just a little less than 3.

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For what its worth:

Your inequality is equivalent to proving $0\le\sum_{m=2}^n {1\over m!}\le \sum_{m=2}^n {1\over 2^{m-1}}<1 .$

Note that for $m\ge2$ we have ${1\over m!}\le {1\over 2^{m-1}}$ and consider the partition of a square of area 1:

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