Upper bound for $|\delta|$ in $1+\delta=\prod_{i=1}^n (1+\varepsilon_i)^{\pm1}$

Question

I am trying to prove the following lemma:

Let $|\varepsilon_i|\leq \varepsilon < \frac{1}{n} $ for all $1\leq i \leq n$ and define $\delta$ with

$$1+\delta=\prod_{i=1}^n (1+\varepsilon_i)^{\pm1}$$

then

$$|\delta| \leq \frac{n\varepsilon}{1-n\varepsilon}$$

The $\pm1$ in the exponent means that the exponent can be $1$ or $-1$ for any $i$ individually.

I am aware that this is a formula that allows approximations of the higher order terms of the product. As I understand it the product term has a leading $1$ and the rest are $\varepsilon_i$ dependant terms.

I have been looking through all my books searching for some useful product inequalities, however the exponent $\pm1$ and the sign of $\varepsilon_i$ rules most of them out, so I cannot use them. This inequality seemed the most promosing to deliver an upper bound:

• Let $a_i \in \mathbb{R_0^+}$ with $\sum_{i=1}^n a_i \leq 1$, then $$\prod_{i=1}^n (1+a_i) \leq 1+2\sum_{i=1}^n a_i $$

However the nonfixed exponent bugs me and does not let me use it.

Are there any theorems or tricks I could use to get to my goal?

I am looking for hints, not for solutions. Thanks.

Answer 1

Your final comments suggests that are already at the solution.

1. Define $\gamma_n = \frac{nu}{1-nu}$ for nonnegative $n$ such that $nu < 1$.

2. Show by direct calculation that $$\gamma_m + \gamma_n + \gamma_m \gamma_n \leq \gamma_{m+n}$$ for nonnegative $m$ and $n$ such that $(m+n)u < 1$.

3. Write $\langle m \rangle $ for any error term of the form $1 + \theta$, where $|\theta| \leq m$.

4. Realize that $$\langle m \rangle \langle n \rangle = \langle m + n \rangle$$ is just the previous inequality.
5. Note that $u \leq \gamma_1$.
6. Show $\frac{1}{1-\epsilon} = \langle 1 \rangle$ for $|\epsilon| \leq u$.

This compact notation, i.e. $\langle m \rangle$ saves you the grief of inventing sensible names and it automatically keeps track of the size of the error terms.