Convergence of $\sum^\infty_{n=0} \frac{\cos(n + \frac{1}{n^2})}{n \cdot \ln(n^2 + 1)}$ Is my idea correct? Convergence:
$$\sum^\infty_{n=0}\frac{\cos \left(n + \frac{1}{n^2} \right)}{n \cdot \ln \left( n^2 + 1 \right)}$$
Edit (I tried to follow the idea written by @Daniel Fischer):
$$\sum_{n=0}^{\infty} \frac{cos(n + \frac{1}{n^2})}{n\cdot ln(n^2 + 1)} = \sum_{n=0}^{\infty} \frac{cos(n)}{n \cdot ln(n^2 + 1)} + \sum_{n=0}^{\infty} \frac{cos(n + \frac{1}{n^2}) - cos(n)}{n \cdot ln(n^2 + 1)}$$
$\sum_{n=0}^{\infty} \frac{cos(n)}{n \cdot ln(n^2 + 1)}$ is convergent because of Dirichlet test and How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?
$\sum_{n=0}^{\infty} \frac{cos(n + \frac{1}{n^2}) - cos(n)}{n \cdot ln(n^2 + 1)} = \sum_{n=0}^{\infty} \frac{-2 \left(sin \left( \frac{2n+\frac{1}{n^2}}{2} \right) \cdot sin \left( \frac{\frac{1}{n^2}}{2} \right) \right)}{n \cdot ln \left(n^2 + 1 \right)} = \sum_{n=0}^{\infty} \frac{-2 \left(sin \left( n+\frac{1}{2n^2} \right) \cdot sin \left( \frac{1}{2n^2} \right) \right)}{n \cdot ln \left(n^2 + 1 \right)}$
Then:
$$\sum_{n=0}^{\infty} \left| \frac{-2 \left(sin \left( n+\frac{1}{2n^2} \right) \cdot sin \left( \frac{1}{2n^2} \right) \right)}{n \cdot ln \left(n^2 + 1 \right)} \right| \leq \sum_{n=0}^{\infty} \left| 2 \cdot \frac{ \frac{1}{2n^2} }{n \cdot ln \left(n^2 + 1 \right)} \right| \leq \sum_{n=0}^{\infty} \left| \frac{ 1 }{n^3 \cdot ln \left(n^2 + 1 \right)} \right| $$
For $n \geq 3$:
$$\sum_{n=0}^{\infty} \left| \frac{ 1 }{n^3 \cdot ln \left(n^2 + 1 \right)} \right| \leq \sum_{n=0}^{\infty} \left| \frac{1}{n^2} \right|$$
Therefore we know that $\sum_{n=0}^{\infty} \frac{cos(n + \frac{1}{n^2}) - cos(n)}{n \cdot ln(n^2 + 1)}$ it is also convergent. Is that correct?

Continuing from the remark made in the comments,
$$\cos\biggl(n + \frac{1}{n^2}\biggr) = \cos n + \biggl(\cos \biggl(n + \frac{1}{n^2}\biggr) - \cos n\biggr).$$
The second term on the right decays like $1/n^2$, since cosine is Lipschitz continuous (prove this using the cosine sum identity). The partial sums of the first term on the right are uniformly bounded; to see this, it is convenient to remember cosine is the real part of the complex exponential,
$$ \cos n =  \operatorname{Re} e^{in}. $$
Thus the partial sum of cosine is just the real part of a nice geometric series,
$$ \sum_{n = 0}^m \cos n = \operatorname{Re} \sum_{n = 0}^m e^{in} = \operatorname{Re} \frac{1 - e^{i(m + 1)}}{1 - e^i}. $$
The right hand side has uniformly bounded modulus in $m$, so we are done.
