What is the sum from n=0 to infinity of e^{-0.02n} ?

The sum from n=0 to infinity of e^{-0.02n} is converges

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sum from n=0 to infinity of e^{-0.02n}

n = 0 \sum \infty e^{- 0.02 n}

converges

Solution steps

n = 0 \sum \infty e^{- 0.02 n}

n \to \infty lim (n (\frac{e ^{- 0.02 n}}{e ^{- 0.02 (n + 1)}} - 1)) = \infty

L > 1,

by the Raabe's test

= converges

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What is the sum from n=0 to infinity of e^{-0.02n} ?
The sum from n=0 to infinity of e^{-0.02n} is converges