Calc6.62

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

The terms of this series go to 0, but does it converge? Using the Cauchy condensation test, it is true that this series and

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

converge or diverge together. It is well known that the terms of 2^{k} increase much more quickly than the terms of k so we know that these terms do not go to 0, but instead increase to \infty and thus this series diverges. We can prove this using the ratio test:

\lim _{{k\rightarrow \infty }}\left|{\frac  {{\frac  {2^{{k+1}}}{k+1}}}{{\frac  {2^{k}}{k}}}}\right|=\lim _{{n\rightarrow \infty }}{\frac  {2k}{k+1}}=2\,

Since this number is greater than 1, by the ratio test, this series diverges. Since this series diverges, by the Cauchy condensation test, our original series also diverges.


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