Calc6.32

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Discuss the convergence or divergence of the series \sum _{{n=0}}^{{\infty }}\left({\frac  {1}{6}}\right)^{{n+2}}.

This is a geometric series with common ratio {\frac  {1}{6}} so it converges. To find the sum, we’ll need to reindex the series.

\sum _{{n=0}}^{{\infty }}\left({\frac  {1}{6}}\right)^{{n+2}}=\sum _{{n=2}}^{{\infty }}\left({\frac  {1}{6}}\right)^{n}={\frac  {1}{1-{\frac  {1}{6}}}}-\left({\frac  {1}{6}}\right)^{0}-\left({\frac  {1}{6}}\right)^{1}={\frac  {6}{5}}-{\frac  {1}{6}}-1={\frac  {1}{30}}\,


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