What is the sum from n=1 to infinity of 9/(e^n) ?

The sum from n=1 to infinity of 9/(e^n) is 9(e/(e-1)-1)

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sum from n=1 to infinity of 9/(e^n)

n = 1 \sum \infty \frac{9}{e ^{n}}

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

Solution steps

n = 1 \sum \infty \frac{9}{e ^{n}}

Apply the constant multiplication rule:

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

= 9 \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}

= 9 (n = 0 \sum \infty \frac{1}{e ^{n}} - \frac{1}{e ^{0}})

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

= 9 (\frac{e}{e - 1} - \frac{1}{e ^{0}})

Simplify

= 9 (\frac{e}{e - 1} - 1)

n = 1 \sum \infty \frac{8 n}{5 n + 1}

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

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

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

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

What is the sum from n=1 to infinity of 9/(e^n) ?
The sum from n=1 to infinity of 9/(e^n) is 9(e/(e-1)-1)