Calc6.37

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Discuss the convergence or divergence of the series \sum _{{n=0}}^{{\infty }}\left({\frac  {1}{3}}\right)^{{2n}}.

This is another one that seems like a geometric series but is not quite in the same form. This is a geometric series though. Remember the formula

a^{{mn}}=\left(a^{m}\right)^{n}\,

So here we have

\sum _{{n=0}}^{{\infty }}\left({\frac  {1}{3}}\right)^{{2n}}=\sum _{{n=0}}^{{\infty }}\left[\left({\frac  {1}{3}}\right)^{2}\right]^{n}=\sum _{{n=0}}^{{\infty }}\left({\frac  {1}{9}}\right)^{{n}}={\frac  {1}{1-{\frac  {1}{9}}}}={\frac  {9}{8}}\,


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