Calc6.65

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Discuss the convergence or divergence of the series \sum _{{k=1}}^{{\infty }}{\frac  {1}{(2^{k})^{n}}}, where n is some real number.

Let us multiply the numerator and denominator of this by 2^{k}. This gives us

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

This is the form

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

and thus converges or diverges together with the series

\sum _{{k=1}}^{{\infty }}{\frac  {1}{k^{{n+1}}}}.

This, we know converges if n>0 and diverges for any other value of n (n\leq 0) by the n-series test. Thus, our original series converges if n>0 and diverges if n\leq 0.


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