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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.

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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.

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By considering logarithmic derivatives and the convexity of the tangent function, it is straightforward to check that: $$\forall x\in(0,1),\quad x^2\leq 2\log(\sec x)\leq x^2\tan 1$$ hence the given series is convergent by comparison with the generalized harmonic series: $$\sum_{n\geq 1}\frac{1}{n^2}=\frac{\pi^2}{6}.$$

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