I am struggling with the problem in the title of this post. I have tried many different methods, but nothing has worked so far. I only managed to derive the first term of the asymptotic expansion:
$f(n)=\sum\limits_{k=1}^{\infty}\frac{(-1)^k}{k}\ln(n^2+k) =\sum\limits_{k=1}^{\infty}\frac{(-1)^k}{k}\ln(n^2) + \sum\limits_{k=1}^{\infty}\frac{(-1)^k}{k}\ln\left(1+\frac{k}{n^2}\right) = -\ln (2) \ln(n^2)+ o(1)$, $\quad n \to +\infty$,
where I used $\ln(n^2+k)=\ln\left( n^2\left(1+\frac{k}{n^2}\right)\right)= \ln(n^2) +\ln \left(1+\frac{k}{n^2}\right)$.
How can I derive the next expansion term?