What is the sum from n=1 to infinity of (7/8)^{n-1} ?

The sum from n=1 to infinity of (7/8)^{n-1} is converges

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sum from n=1 to infinity of (7/8)^{n-1}

n = 1 \sum \infty (\frac{7}{8})^{n - 1}

converges

Solution steps

n = 1 \sum \infty (\frac{7}{8})^{n - 1}

Apply Series Ratio Test:

converges

= converges

\frac{d}{d x} (e^{\frac{x}{x - 8}})

\int (9 x^{2} - 8 x) e^{2 x} d x

y^{^{''}} - 6 y + 9 y = 0

\frac{d y}{d x} = \frac{( 2 x ^{2} + 3 ) ( 3 x + 2 )}{7 x - 3}

(x + 1) \frac{d y}{d x} = x + 6

What is the sum from n=1 to infinity of (7/8)^{n-1} ?
The sum from n=1 to infinity of (7/8)^{n-1} is converges