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

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

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

n = 0 \sum \infty \frac{2}{7 ^{n}}

\frac{7}{3}

Solution steps

n = 0 \sum \infty \frac{2}{7 ^{n}}

Apply the constant multiplication rule:

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

= 2 \cdot n = 0 \sum \infty \frac{1}{7 ^{n}}

Apply Series Geometric Test:

\frac{7}{6}

= 2 \cdot \frac{7}{6}

Simplify

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

= \frac{7}{3}

x \to 2 - lim (arcsin (x - 1))

x \to π lim (\frac{1}{x - π})

f (x) = (6 x + 1)^{2}

\frac{d}{d x} (ln (8 x))

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