Calc6.36

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Discuss the convergence or divergence of the series \sum _{{n=0}}^{{\infty }}{\frac  {3^{{n+1}}}{7^{n}}}.

This is not quite in the same form as our formula but it is a geometric series in a slight disguise.

\sum _{{n=0}}^{{\infty }}{\frac  {3^{{n+1}}}{7^{n}}}=3\sum _{{n=0}}^{{\infty }}\left({\frac  {3}{7}}\right)^{n}=3{\frac  {1}{1-{\frac  {3}{7}}}}={\frac  {21}{4}}\,

So, this geometric series, with common ratio {\frac  {3}{7}}, converges to {\frac  {21}{4}}.


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