<s>[INST] Series convergence study using big O? Can someone explain why the following expression:
$\sum_{n=1}^{\infty} (-1)^{n}(\frac{1}{n})(\frac{\frac{7}{12}+O(\frac{1}{n^{2}})}{-\frac{1}{6}+O(\frac{1}{n^{2}})})$
can be rewritten as:
$\sum_{n=1}^{\infty} (-1)^{n}(\frac{1}{n})(\frac{-7}{2}+O(\frac{1}{n^{2}}))(1+O(\frac{1}{n^{2}}))$
And why we are allowed to split this last expression of the series into a convergent part and an absolutely convergent one?
Still trying to wrap my head around big O notation. Thanks in advance.
 [/INST] One may recall that, as $x \to 0$, by the Taylor series expansion we have
$$
\frac1{1-x}=1+x+O(x^2)
$$ giving, as $n \to \infty$,
$$
\frac{1}{-\frac{1}{6}+O(\frac{1}{n^2})}=(-6)\times\frac{1}{1-O(\frac{1}{n^2})}=(-6)\times\left( 1+O\left(\frac{1}{n^2}\right)\right)
$$ and
$$
\frac{\frac{7}{12}+O(\frac{1}{n^{2}})}{-\frac{1}{6}+O(\frac{1}{n^{2}})}=\left(\frac{7}{12}+O\left(\frac{1}{n^{2}}\right)\right)\times(-6)\times\left( 1+O\left(\frac{1}{n^2}\right)\right)=-\frac72+O\left(\frac{1}{n^2}\right)
$$ as wanted. 
The inital series converges being the sum of two convergent series.
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