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?
 A: Alternative way using Taylor series:
$$\ln\left(\cos \frac{1}{n}\right) = \ln\left(1-O\left(\frac{1}{n^2}\right) \right) = O\left(\frac{1}{n^2}\right)$$
A: For every $n\geqslant 1$, $\cos \dfrac{1}{n}$ gets closer and closer to $1$.
That is,
$$\cos \dfrac{1}{n}\approx 1-\dfrac{1}{2n^2}+\dfrac{1}{24n^4}-\dfrac{1}{720n^6}$$ and for large enough $n$, the larger terms go to zero, making the approximation more accurate; as $n$ approaches $\infty$, our polynomial, and the cosine, go to $1$. Thus in the sum
$$\sum_{n\geqslant 1}\ln\left(\cos\dfrac{1}{n}\right)$$ the terms
$$\ln\left(\cos\dfrac{1}{n}\right)\longrightarrow 0,$$ since $\ln 1 = 0$. The approximate value of $$\ln\cos \dfrac{1}{n}\approx -\dfrac{1}{2n^2} - \dfrac{1}{12n^4} - \dfrac{1}{45n^6} - \dfrac{17 }{2520n^8}$$ and the terms go to zero for sufficiently large $n$. Therefore your series converges and is (for fun) approximately
$$-\left[\frac{1}{2}\zeta(2)+\dfrac{1}{12}\zeta (4)+\dfrac{1}{45}\zeta(6)+\dfrac{17}{2520}\zeta(8)\right]$$ though this agrees only to two decimal places with the correct figure $-0.945369$.
A: 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
$$\left(
\begin{array}{cc}
 1 & -0.9295780173 \\
 2 & -0.9433749128 \\
 3 & -0.9449845778 \\
 4 & -0.9452801699 \\
 5 & -0.9453463832 \\
 6 & -0.9453628819 \\
 7 & -0.9453672911 \\
 8 & -0.9453685316 \\
 9 & -0.9453688948 \\
 10 & -0.9453690046 
\end{array}
\right)$$
