What is the sum from n=0 to infinity of 9e^{-9n} ?

The sum from n=0 to infinity of 9e^{-9n} is converges

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sum from n=0 to infinity of 9e^{-9n}

n = 0 \sum \infty 9 e^{- 9 n}

converges

Solution steps

n = 0 \sum \infty 9 e^{- 9 n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 9 \cdot n = 0 \sum \infty e^{- 9 n}

Apply Series Integral Test:

converges

= 9 converges

= converges

n = 0 \sum \infty 10 (\frac{3}{4})^{n}

n = 0 \sum \infty \frac{1}{2 ^{3 n}}

n = 0 \sum \infty \frac{5}{1 0 ^{n}}

k = 1 \sum \infty \frac{k}{ln ( k )}

n = 1 \sum \infty \frac{7 ^{n}}{n ! n}

What is the sum from n=0 to infinity of 9e^{-9n} ?
The sum from n=0 to infinity of 9e^{-9n} is converges