What is the sum from n=1 to infinity of 3/(e^n) ?

The sum from n=1 to infinity of 3/(e^n) is 3(e/(e-1)-1)

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sum from n=1 to infinity of 3/(e^n)

n = 1 \sum \infty \frac{3}{e ^{n}}

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

Solution steps

n = 1 \sum \infty \frac{3}{e ^{n}}

Apply the constant multiplication rule:

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

= 3 \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}

= 3 (n = 0 \sum \infty \frac{1}{e ^{n}} - \frac{1}{e ^{0}})

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

= 3 (\frac{e}{e - 1} - \frac{1}{e ^{0}})

Simplify

= 3 (\frac{e}{e - 1} - 1)

\int \frac{1}{x ^{2} + 2 x + 10} d x

\int x (2 + x^{2})^{3} d x

\frac{- a - b}{( s + 1 )}

\frac{2}{3} x^{\frac{3}{2}}

\frac{3}{s ( s + 2 )}

What is the sum from n=1 to infinity of 3/(e^n) ?
The sum from n=1 to infinity of 3/(e^n) is 3(e/(e-1)-1)