I would appreciate help in understanding how:
$$\log \left(\frac{1}{s - 1} - O(1)\right) = \log \left(\frac{1}{s - 1}\right) + O(1)\text{ as }s \rightarrow 1^+$$
I thought of perhaps a Taylor series for $\log(x - 1)$, but that would have an $O(x^2)$ term.
Also I would appreciate any reference or tutorial recommendations to get a good understanding of big-$O$ notation in general.
Thanks very much.
On
$$\begin{align}\log\left(\frac{1}{s-1}+\mathcal O(1)\right)&=\log\left(\frac{1}{s-1}\left(1+\mathcal O(s-1)\right)\right)=\log\left(\frac{1}{s-1}\right)+\log\left(1+\mathcal O(s-1)\right)\\ \notag &=\log\left(\frac{1}{s-1}\right)+\mathcal O(s-1) \\ \notag &=\log\left(\frac{1}{s-1}\right)+\mathcal O(1)\end{align}$$
the third equality holding since $$\log(1+x)=\mathcal O(x),\quad x\to0.$$
Some relevant notes here (http://www.ma.hw.ac.uk/~simonm/ae.pdf).
$f(s)=O(1)$ as $s\rightarrow1^+$ means that there is a constant $C$ such that $|f(s)|<C$ for $s$ close to $1^+$.
One thing to always keep in mind is that equalities between things with $O$'s or $o$'s are not really symmetric equalities. This means that it is probably safe to prove this 'in both directions'.
The left hand side is talking about a function $f(s)$ such that there is a constant $C$ such that $|e^{f(s)}-\frac{1}{s-1}|<C$ for $s$ close to $1^+$.
The right hand side is talking about a function $g(s)$ such that there is a constant $D$ such that $|g(s)-\log(\frac{1}{s-1})|<D$.
Let us check that the function on the left $f$ satisfies the description on the right. According to its description, we can write $f(s)$ as $\log(\frac{1}{s-1}+h(s))$, where $h$ is bounded as $s\rightarrow 1^+$. We subtract $\log(\frac{1}{s-1})$ and get $\log(1+(s-1)h(s))$. Since $h$ is bounded and $(s-1)\rightarrow 0$ then this subtraction is bounded. This was the description on the right.
Now, we do the other direction. Let us check that $g$ satisfies the description on the left. From what we know of $g$ we can write it as $\log(\frac{1}{s-1})+a(s)$ with $a$ bounded. We now compute $e^{\log(1/(s-1))+a(s)}-\frac{1}{s-1}=\frac{e^{a(s)}}{(s-1)}-\frac{1}{(s-1)}=\frac{e^{a(s)}-1}{(s-1)}$ ... oops.
I guess that is why this is not really a symmetric equality.
So, reading from left to right the equality is true. Reading from right to left, it is not.