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

The sum from n=0 to infinity of 3/(3^n) is 9/2

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

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

\frac{9}{2}

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Geometric Test:

\frac{3}{2}

= 3 \cdot \frac{3}{2}

Simplify

3 \cdot \frac{3}{2} : \frac{9}{2}

= \frac{9}{2}

\frac{x ^{2}}{x ^{2} - 9}

x \to 0 lim (\frac{( 3 e ^{x} - 3 )}{x})

\int + (x) d x

\int 4 tan^{4} (x se) c^{6} x d x

x \to - \infty lim ((1 + \frac{1}{x})^{2 x})

What is the sum from n=0 to infinity of 3/(3^n) ?
The sum from n=0 to infinity of 3/(3^n) is 9/2