Calc6.64

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

In this problem, we will use the Cauchy condensation test in the opposite direction that we have been using it. This series is already in the form

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

which we get by a bit of simplification, or perhaps complication.

\sum _{{k=1}}^{{\infty }}{\frac  {1}{{\sqrt  {2^{k}}}}}=\sum _{{k=1}}^{{\infty }}2^{k}{\frac  {1}{(2^{k})^{{{\frac  {3}{2}}}}}}

Thus, this series converges or diverges with the series

\sum _{{k=1}}^{{\infty }}{\frac  {1}{k^{{{\frac  {3}{2}}}}}}.

This series converges because it is a p-series with p>1. Thus, our original series also converges.


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