What is the sum from n=0 to infinity of 6(1/7)^n ?

The sum from n=0 to infinity of 6(1/7)^n is 7

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sum from n=0 to infinity of 6(1/7)^n

n = 0 \sum \infty 6 (\frac{1}{7})^{n}

7

Solution steps

n = 0 \sum \infty 6 (\frac{1}{7})^{n}

Apply the constant multiplication rule:

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

= 6 \cdot n = 0 \sum \infty (\frac{1}{7})^{n}

Apply Series Geometric Test:

\frac{7}{6}

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

Simplify

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

= 7

x \to 2 - lim (\frac{x + 2}{x - 4})

n = - \infty \sum \infty 7^{2}

x \to 2 lim (\frac{3 x}{x - 4})

y^{^{''}} + 2 x = 0

\frac{\partial}{\partial x} (2 x^{2} y - x + 3 y^{3})

What is the sum from n=0 to infinity of 6(1/7)^n ?
The sum from n=0 to infinity of 6(1/7)^n is 7