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

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

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

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

7

Solution steps

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

Apply Series Geometric Test:

7

= 7

\frac{d y}{d x} + 12 y = 4

(2 x y^{2} + 2 y) d x + (2 x^{2} y + 2 x) d y = 0, y (1) = 2

y^{^{''}} - 5 y^{^{'}} + 6 y = - sin (5 t)

\frac{d y}{d x} = \frac{1}{e ^{y} - x}

\frac{d y}{d x}, \frac{6 x}{y ^{7}} = 4

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