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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. – André Nicolas Nov 14 '11 at 6:08
up vote 4 down vote accepted

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:

enter image description here

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