FiessyFrimatsd0

2022-01-25

Prove that $\sum _{n=1}^{\mathrm{\infty }}\mathrm{log}\mathrm{cos}\left(\frac{1}{n}\right)$ converges absolutely.

spelkw

Expert

As $0<\mathrm{cos}\frac{1}{n}<1$, we have $|\mathrm{log}\mathrm{cos}\frac{1}{n}|=\mathrm{log}\left(\frac{1}{\mathrm{cos}\frac{1}{n}}\right)$
Now $\mathrm{cos}\frac{1}{n}=1-\frac{1}{2{n}^{2}}+o\left(\frac{1}{{n}^{2}}\right)$, so
$\frac{1}{\mathrm{cos}\frac{1}{n}}=\frac{1}{1-\frac{1}{2{n}^{2}}+o\left(\frac{1}{{n}^{2}}\right)}=1+\frac{1}{2{n}^{2}}+o\left(\frac{1}{{n}^{2}}\right){\sim }_{\mathrm{\infty }}1+\frac{1}{2{n}^{2}}$
and ultimately
$|\mathrm{log}\mathrm{cos}\frac{1}{n}|{\sim }_{\mathrm{\infty }}\mathrm{log}\left(1+\frac{1}{2{n}^{2}}\right){\sim }_{\mathrm{\infty }}\frac{1}{2{n}^{2}}$
which converges.

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