What is the sum from n=1 to infinity of 6e^{-n} ?

The sum from n=1 to infinity of 6e^{-n} is 6(e/(e-1)-1)

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sum from n=1 to infinity of 6e^{-n}

n = 1 \sum \infty 6 e^{- n}

6 (\frac{e}{e - 1} - 1)

Solution steps

n = 1 \sum \infty 6 e^{- n}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty e^{- n}

Simplify

e^{- n} : \frac{1}{e ^{n}}

= 6 \cdot n = 1 \sum \infty \frac{1}{e ^{n}}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= 6 (n = 0 \sum \infty \frac{1}{e ^{n}} - \frac{1}{e ^{0}})

n = 0 \sum \infty \frac{1}{e ^{n}} = \frac{e}{e - 1}

= 6 (\frac{e}{e - 1} - \frac{1}{e ^{0}})

Simplify

= 6 (\frac{e}{e - 1} - 1)

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

k = 5 \sum \infty 3 (\frac{1}{3})^{3 k}

n = 2 \sum \infty (\frac{1}{5})^{2 n}

k = 0 \sum \infty (0.3)^{k}

n = 1 \sum \infty (cos (2))^{n}

What is the sum from n=1 to infinity of 6e^{-n} ?
The sum from n=1 to infinity of 6e^{-n} is 6(e/(e-1)-1)