Calc6.29

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Discuss the convergence or divergence of the series \sum _{{n=0}}^{{\infty }}\left({\frac  {1}{4}}\right)^{n}.

This is a geometric series with common ratio {\frac  {1}{4}} so this series converges and the sum is

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


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