Show that $\frac{1}{e^\gamma \text{log }x + O(1)} = \frac{1}{e^\gamma\text{log }x} + O\left(\frac{1}{(\text{log }x)^2}\right)$

Question

Show that $\frac{1}{e^\gamma \text{log }x + O(1)} = \frac{1}{e^\gamma\text{log }x} + O\left(\frac{1}{(\text{log }x)^2}\right)$

I'm using one of Merten's estimates in a proof, the one that states

\begin{align} \prod\limits_{p \leq x}\left(1-\frac{1}{p}\right)^{-1}=e^\gamma\text{log }x +O(1) \end{align}

To find $\prod\limits_{p \leq x}\left(1-\frac{1}{p}\right) = \frac{1}{e^\gamma\text{log }x} + O\left(\frac{1}{(\operatorname{log}x)^2}\right)$ but I am loss at how to show this, any help is appreciated greatly.

Answer 1

Would an answer like this work?:

\begin{align} \frac{1}{e^\gamma\text{log }x + O(1)} &= \frac{1+O(\text{log }x)}{e^\gamma\text{log }x + O((\text{log }x)^2)}\\ &=\frac{1}{e^\gamma\text{log }x}\left(\frac{1+O(\text{log }x)}{1+O(\text{log }x)}\right)\\ &=\frac{1}{e^\gamma\text{log }x}(1+O(\text{log }x)) \end{align}

Answer 2

You might want to work with logarithms. Since $\log x+\mathcal O(1)=\log x\{1+\mathcal O(1/\log x)\}$, we have

$$ -\log\prod_{p\le x}\left(1-\frac1p\right)=\gamma+\log\log x+\mathcal O\left(1\over\log x\right) $$

Now multiply both sides by $-1$ and exponentiate, so we get

$$ \prod_{p\le x}\left(1-\frac1p\right)={e^{-\gamma}\over\log x}\left\{1+\mathcal O\left(1\over\log x\right)\right\} $$

Finally, expanding the parenthesis, we get the desired result.