Calc6.2

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Discuss the convergence of \sum _{{n=0}}^{{\infty }}e^{{-3n}}.

Again, use the integral test. Notice that the integral test states the series starts at n=1 and this series starts at n=0. This does not change the fact that the integral test works. One extra term, or a million for that matter, will not change the convergence or divergence of a series. The one thing you must test is to make sure the function is continuous on the entire interval. For example, {\frac  {1}{n}} is undefined when n=0 so starting such a series at 0 would not work.

\int _{{0}}^{{\infty }}e^{{-3x}}\,dx=-{\frac  {1}{3}}e^{{-3x}}{\bigg |}_{{0}}^{{\infty }}=0-\left(-{\frac  {1}{3}}e^{{-3\times 0}}\right)={\frac  {1}{3}}\,

This integral converges to a real number. Since this function is continuous, positive, and decreasing, by the integral test, our series also converges.


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