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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.

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    $\begingroup$ Hint: For large $y$, $$\frac{1}{y+1} = \frac{1}{y} \frac{1}{1+\frac{1}{y}}.$$ Now expand the second factor, $1/(1+1/y)$, as a geometric series. $\endgroup$ – Antonio Vargas Oct 19 '14 at 2:22
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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}

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  • $\begingroup$ It's not clear to me how you get the first equality. Further, the last expression doesn't yield the correct answer. $\endgroup$ – Antonio Vargas Oct 19 '14 at 2:19

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