Determine whether the series $\sum_{n=1}^\infty \ln\left(\cos \frac{1}{n}\right)$ is convergent. Determine whether the series $\sum_{n=1}^\infty \ln\left(\cos \frac{1}{n}\right)$ is convergent.
Attempt:
Notice that the $p-$ series $\sum_{n=1}^\infty \frac{1}{n^2}$ with $p=2$ is convergent. Then, we have
\begin{align*}
\lim\limits_{n \to \infty} \frac{\ln\left(\cos \frac{1}{n} \right)}{\frac{1}{n^2}} &= -\frac{1}{2} \cdot \lim\limits_{n \to \infty} \frac{\sin \frac{1}{n}}{\frac{1}{n}} \cdot \frac{1}{\cos \frac{1}{n}} \\
&= -\frac{1}{2}.
\end{align*}
Hence, the limit is equal to $-\frac{1}{2} \ne 0$. Since the $p-$ series is convergent with $p=2$, by the limit comparison test, the series is convergent.
Am I true?

Just for the fun.
In the same spirit as @Alexander Conrad, we have the infinite series representation
$$\log \left(\cos \left(\frac{1}{n}\right)\right)=\sum_{k=1}^\infty (-1)^k\frac{ 2^{2 k-3} (E_{2 k-1}(1)-E_{2 k-1}(0)) }{k (2 k-1)!}\,n^{-2 k}$$ where appear Euler polynomials.
Truncated to any order, we can transform the Taylor series into a $[2k,2]$ Padé approximant $P_k$.
For  example
$$P_2=-\frac {3(10n^2-1) } {4n^2(15n^2-4) }=-\frac{3}{16 n^2}-\frac{75}{16 \left(15 n^2-4\right)}$$
$$\sum_{n=1}^\infty P_2=-\frac{1}{128} \left(75+4 \pi ^2-10 \sqrt{15} \pi  \cot \left(\frac{2 \pi}{\sqrt{15}}\right)\right)\sim -0.943375$$ Repeating for various values of $k$, we have more and more accurate values as shown below
$$\left(
\begin{array}{cc}
 1 & -0.9295780173 \\
 2 & -0.9433749128 \\
 3 & -0.9449845778 \\
 4 & -0.9452801699 \\
 5 & -0.9453463832 \\
 6 & -0.9453628819 \\
 7 & -0.9453672911 \\
 8 & -0.9453685316 \\
 9 & -0.9453688948 \\
 10 & -0.9453690046 
\end{array}
\right)$$
