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 3/2

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

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

\frac{3}{2}

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}}

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}{3 ^{n}} - \frac{1}{3 ^{0}})

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

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

Simplify

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

= \frac{3}{2}

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

\int \frac{t ^{2}}{t ^{3} + 7} d t

\frac{d y}{d x} + 8 y - 8 e^{- 5 x} = 0, y (0) = \frac{5}{3}

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

y = \frac{6}{x - 1}, (3, 3)

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 3/2