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Question

Test the series for convergence.

n=1en\sum_{n=1}^{\infty} e^{-n}

Solution

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The given series

n=1en=n=11en=n=1(1e)n\sum_{n=1}^\infty e^{-n}=\sum_{n=1}^\infty \frac{1}{e^n}=\sum_{n=1}^\infty (\frac{1}{e})^n

corresponds to geometric series with the factor r=1er=\frac{1}{e}. Recall: e2,7183e\approx 2,7183. Since r<1|r|<1, the given geometric series converges.

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