calculus - Does the series $sum_{i=1}^{infty}log(sec(frac1{n}))$ converge?

Does the series $\sum_{i=1}^{\infty}\log(\sec(\frac1{n}))$ converge?

My try:As $n$ approaches zero, $\sec(\frac 1n)$ gets close to $\frac 1{1-0.5\frac 1{n^2}}=\frac{2n^2}{2n^2-1}=1+\frac {1}{2n^2-1}$. Since $\frac 1{2n^2-1}$ gets close to zero, approximately $\log(\sec(\frac 1n))=\frac 1{2n^2-1}$. Then compute the limit: $\lim_{n\to \infty}\log(\sec(\frac 1n))/\frac 1{2n^2-1}=\lim_{n\to{\infty}}\tan(\frac 1n)\frac{(2n^2-1)^2}{4n^3}$, so the limit diverges and limit comparison test can bot be applied. Vanishing test is inconclusive since the term obviously goes to zero. Ratio test does not help either. I hope someone could help.

Answer

We have $a_n = -\log(\cos(1/n)) = -\frac12 \log(\cos^{2}(1/n)) = -\frac12 \log(1-\sin^{2}(1/n))$. We have
$$\lim_{n \to \infty} \dfrac{a_n}{1/n^2} = -\frac12 \cdot \lim_{n \to \infty} n^2 \log(1-\sin^2(1/n)) = \frac12$$
Hence, by limit comparison test the series converges since $\sum_n \frac1{n^2}$ converges.