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Discuss the convergence or divergence of the series \sum _{{k=3}}^{{\infty }}{\frac  {1}{\ln(\ln k)}}\cos \left({\frac  {k\pi }{3}}\right).

We know {\frac  {1}{\ln(\ln k)}}\rightarrow 0 as k\rightarrow \infty , though at an extremely slow rate. So, if we can show that the sequence of partial sums

\{s_{n}\}=\left\{\sum _{{k=3}}^{{n}}\cos \left({\frac  {k\pi }{3}}\right)\right\}

is bounded, then we can conclude by Dirichlet's test that our given series converges.

\cos 0=1\,

\cos {\frac  {\pi }{3}}={\frac  {1}{2}}

\cos {\frac  {2\pi }{3}}=-{\frac  {1}{2}}

\cos \pi =-1\,

\cos {\frac  {4\pi }{3}}=-{\frac  {1}{2}}

\cos {\frac  {5\pi }{3}}={\frac  {1}{2}}

And, at this point, we get repeat since \cos(2\pi )=\cos 0. Thus, our sequence of partial sums is (notice k starts at 3)

\{s_{n}\}=\{-{\frac  {1}{2}},-{\frac  {3}{2}},-2,-{\frac  {3}{2}},-{\frac  {1}{2}},0,-{\frac  {1}{2}},-{\frac  {3}{2}},-2,-{\frac  {3}{2}},-{\frac  {1}{2}},0,...\}

Thus, this sequence is bounded and, by Dirichlet's test, this series converges.

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