I have the following series
$$\sum_{k=1}^{\infty} \log\left(1+\frac{1}{k^2}\right)$$
This series should converge but when I apply the Direct comparison test it diverges
$$\left|\sum_{k=1}^{\infty} \log(1+\frac{1}{k^2})\right| \le \sum_{k=1}^{\infty} \log\left|1+\frac{1}{k^2}\right| \le \sum_{k=1}^{\infty} 1+\frac{1}{k^2} = \sum_{k=1}^{\infty} 1 + \sum_{k=1}^{\infty}\frac{1}{k^2}$$
so we know that $\sum_{k=1}^{\infty} \left(1+\frac{1}{k^2}\right) = \sum_{k=1}^{\infty} 1 + \sum_{k=1}^{\infty}\frac{1}{k^2}$ diverges because $\sum 1$ diverges, so the series should diverges.
what am I doing wrong?
Should it be in this way :
$$\left|\sum_{k=1}^{\infty} \log(1+\frac{1}{k^2})\right| \le \sum_{k=1}^{\infty} \log\left|1+\frac{1}{k^2}\right| \le \sum_{k=1}^{\infty} \frac{1}{k^2}$$
so it will converges because $\sum \frac{1}{k^2}$ converges ? if yes, why do we ignore $\sum 1$