What is the sum from n=0 to infinity of 3/(7^n) ?

The sum from n=0 to infinity of 3/(7^n) is 7/2

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sum from n=0 to infinity of 3/(7^n)

n = 0 \sum \infty \frac{3}{7 ^{n}}

\frac{7}{2}

Solution steps

n = 0 \sum \infty \frac{3}{7 ^{n}}

Apply the constant multiplication rule:

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

= 3 \cdot n = 0 \sum \infty \frac{1}{7 ^{n}}

Apply Series Geometric Test:

\frac{7}{6}

= 3 \cdot \frac{7}{6}

Simplify

3 \cdot \frac{7}{6} : \frac{7}{2}

= \frac{7}{2}

\frac{d}{d x} ((7 x + 1) (x^{4} - x^{3} - 9 x))

y^{^{''}} + 4 y^{^{'}} + 3 y = 2 x

\frac{d}{d x} (\frac{e ^{x}}{2 x ^{2}})

\int 1 - sin^{2} (x) d x

\frac{d}{d x} (16 + x)

What is the sum from n=0 to infinity of 3/(7^n) ?
The sum from n=0 to infinity of 3/(7^n) is 7/2