# Calc6.64

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.