What is the sum from n=1 to infinity of 5e^{-0.5n} ?

The sum from n=1 to infinity of 5e^{-0.5n} is converges

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sum from n=1 to infinity of 5e^{-0.5n}

n = 1 \sum \infty 5 e^{- 0.5 n}

converges

Solution steps

n = 1 \sum \infty 5 e^{- 0.5 n}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty e^{- 0.5 n}

Apply Series Ratio Test:

converges

= 5 converges

= converges

n = 0 \sum \infty \frac{7}{n + 3}

n = 1 \sum \infty \frac{2 n ^{3}}{n ^{4}}

n = 2 \sum \infty (1 + \frac{1}{2})^{- n}

n = 1 \sum \infty \frac{- 1}{n ^{2}}

n = 1 \sum \infty \frac{2 n + 3}{n + 1}

What is the sum from n=1 to infinity of 5e^{-0.5n} ?
The sum from n=1 to infinity of 5e^{-0.5n} is converges