Calc6.61

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Discuss the convergence or divergence of the series \sum _{{k=1}}^{{\infty }}{\frac  {1}{k}}.

By the Cauchy condensation test, we know that this series and the series

\sum _{{k=1}}^{{\infty }}2^{k}{\frac  {1}{2^{k}}}

either both converge or both diverge. Looking at this new series, it is clear to see that it diverges.

\sum _{{k=1}}^{{\infty }}2^{k}{\frac  {1}{2^{k}}}=\sum _{{k=1}}^{{\infty }}1

The terms do not even go to 0 in this series, so by the nth term test, it diverges. Thus, by the Cauchy condensation test, it is also true that our series diverges.


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