What is the sum from n=1 to infinity of 5/(e^n) ?

The sum from n=1 to infinity of 5/(e^n) is 5(e/(e-1)-1)

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sum from n=1 to infinity of 5/(e^n)

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

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

Solution steps

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

Apply the constant multiplication rule:

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

= 5 \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}

= 5 (n = 0 \sum \infty \frac{1}{e ^{n}} - \frac{1}{e ^{0}})

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

= 5 (\frac{e}{e - 1} - \frac{1}{e ^{0}})

Simplify

= 5 (\frac{e}{e - 1} - 1)

n = 1 \sum \infty (sin (400))^{n}

n = 0 \sum \infty \frac{n !}{n + 1}

n = 1 \sum \infty \frac{1}{7 n ^{3} + 3}

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

k = 1 \sum \infty \frac{k ^{k}}{2 ^{k}}

What is the sum from n=1 to infinity of 5/(e^n) ?
The sum from n=1 to infinity of 5/(e^n) is 5(e/(e-1)-1)