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

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

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

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

3

Solution steps

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

Apply the constant multiplication rule:

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

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

Simplify

= 6 \cdot n = 1 \sum \infty \frac{1}{3 ^{n}}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= 6 (n = 0 \sum \infty \frac{1}{3 ^{n}} - \frac{1}{3 ^{0}})

n = 0 \sum \infty \frac{1}{3 ^{n}} = \frac{3}{2}

= 6 (\frac{3}{2} - \frac{1}{3 ^{0}})

Simplify

6 (\frac{3}{2} - \frac{1}{3 ^{0}}) : 3

= 3

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

n = 1 \sum \infty \frac{n ^{4}}{4 n !}

x = 1 \sum \infty \frac{x}{x ( x + 1 )}

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

k = 1 \sum \infty k e^{- 16 k}

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