# Calc6.2

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