Check if $\sum_{n=1}^{\infty}\left|\frac{\ln(n+1)}{\ln(n^2+1)}-\frac{1}{2}\right|$ converges This is the series:
$$\sum_{n=1}^{\infty}\left|\frac{\ln(n+1)}{\ln(n^2+1)}-\frac{1}{2}\right|$$
We can drop the absolute value, because;
$$\frac{\ln(n+1)}{\ln(n^2+1)} > \frac{1}{2}$$
I've tried using inequalities below to form comparison tests, but to no avail.
$$\frac{n}{n+1} < \ln(n+1) < n$$
$$\frac{1}{n^2} < \frac{1}{\ln(n^2+1)} < \frac{n^2+1}{n^2}$$
I also tried d'Alembert's ratio test, but it resulted in limit of 1, which is inconclusive. I have no clues on how to approach this.

One has
\begin{align*}
\frac{\ln(n+1)}{\ln(n^2+1)} -\frac{1}{2} &= \frac{\ln(n)+\ln\left(1+\frac{1}{n}\right)}{\ln(n^2)+\ln\left(1+\frac{1}{n^2}\right)} -\frac{1}{2} \\
&= \frac{\ln(n)+ \frac{1}{n}+ o\left( \frac{1}{n}\right)}{2\ln(n)+  o\left( \frac{1}{n}\right)} -\frac{1}{2} \\
&= \frac{ \frac{1}{n} + o\left( \frac{1}{n}\right)}{2\ln(n) + o\left( \frac{1}{n}\right)} \sim \frac{1}{2n\ln(n)}
\end{align*}
This is the general term of a divergent series, hence by comparison, the original series diverges.
