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

This converges because it is a geometric series with common ratio {\frac  {2}{3}}. Be careful when computing the sum. Our formula is for a sum starting at n=0 and here we start with n=3. Thus

\sum _{{n=3}}^{{\infty }}\left({\frac  {2}{3}}\right)^{n}={\frac  {1}{1-{\frac  {2}{3}}}}-\left({\frac  {2}{3}}\right)^{0}-\left({\frac  {2}{3}}\right)^{1}-\left({\frac  {2}{3}}\right)^{2}=3-1-{\frac  {2}{3}}-{\frac  {4}{9}}={\frac  {8}{9}}\,

So, this geometric series converges to the sum {\frac  {8}{9}}.

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