Deriving the asymptotic estimate (9.62) in Concrete Mathematics

We have $\enspace\displaystyle n g_n = \frac{1}{n}\sum\limits_{k\ge 0}g_k - \frac{1}{n}\sum\limits_{k\ge n}g_k + \frac{1}{n}\sum\limits_{0<k<n}\frac{k}{n-k}g_k $

with $\enspace\displaystyle g_n = \frac{G(1)}{n^2} + O\left(\frac{\ln n}{n}\right)^3 \enspace$ and $\enspace\displaystyle G(1)=e^{\zeta(2)}\,$ .

First part: $\enspace\displaystyle \sum\limits_{k\ge 0}g_k = G(1) $

With $\enspace\displaystyle O\left(\frac{G(1)}{n^2} + O\left(\frac{\ln n}{n}\right)^3\right)= O\left(\frac{1}{n^2}\right)\enspace$ we get

for the second part $\enspace\displaystyle \sum\limits_{k\ge n}g_k = O\left(\sum\limits_{k\ge n}\frac{1}{k^2}\right)=O\left(\frac{1}{n}\right)\,$ ,

where the last step comes from $\enspace\displaystyle \lim\limits_{n\to\infty}n^x\sum\limits_{k\ge n} \frac{1}{k^{1+x}}=\frac{1}{x}\enspace$ for $\enspace x>0\,$ .

The third part is $\enspace\displaystyle \sum\limits_{0<k<n}\frac{k}{n-k}g_k = O\left(\sum\limits_{0<k<n}\frac{1}{k(n-k)}\right)=O\left(\frac{\ln n}{n}\right) \,$ .

It follows: $\enspace\displaystyle g_n =\frac{1}{n}\left(\frac{G(1)}{n} - \frac{1}{n} O\left(\frac{1}{n}\right) + \frac{1}{n} O\left(\frac{\ln n}{n}\right)\right) = \frac{G(1)}{n^2} + O\left(\frac{\ln n}{n^3}\right)$