Calc6.63

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Discuss the convergence or divergence of the series \sum _{{k=1}}^{{\infty }}{\frac  {1}{k^{2}}}.

By now, you probably know that this series converges since it is a p-series with p=2. However, let us look at this series again using the Cauchy condensation test. By this test, we know that our series and the series

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

converge or diverge together. It is easy to show that this series converges by simplifying it.

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

This is just a geometric series with common ratio {\frac  {1}{2}} and first term {\frac  {1}{2}} which is well known to converge to 1. Thus, this series converges and therefore the original series converges by the Cauchy condensation test.


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