What is the sum from n=0 to infinity of 3/(e^n) ?

The sum from n=0 to infinity of 3/(e^n) is 3*e/(e-1)

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sum from n=0 to infinity of 3/(e^n)

n = 0 \sum \infty \frac{3}{e ^{n}}

3 \cdot \frac{e}{e - 1}

Solution steps

n = 0 \sum \infty \frac{3}{e ^{n}}

Apply the constant multiplication rule:

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

= 3 \cdot n = 0 \sum \infty \frac{1}{e ^{n}}

Apply Series Geometric Test:

\frac{e}{e - 1}

= 3 \cdot \frac{e}{e - 1}

n = 0 \sum \infty π^{- n}

n = 1 \sum \infty \frac{n}{5 n ^{2} + 3}

n = 0 \sum \infty \frac{2}{n}

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

n = 1 \sum \infty 2 + e^{- n}

What is the sum from n=0 to infinity of 3/(e^n) ?
The sum from n=0 to infinity of 3/(e^n) is 3*e/(e-1)