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

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

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sum from n=0 to infinity of 1/(3e^n)

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

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

Decimal

0.52732 \dots

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Geometric Test:

\frac{e}{e - 1}

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

Simplify

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

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

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