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

The sum from n=7 to infinity of (6/7)^{-n} is diverges

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sum from n=7 to infinity of (6/7)^{-n}

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

y = x^{6}

\frac{5 x}{2}, - \frac{1}{2} x + 3, [0, 1]

\frac{( 2 s + 2 )}{s + 10}

\int \frac{1}{4 u ^{2} + 23} d u

\frac{d}{d x} ((2 + x) x)

What is the sum from n=7 to infinity of (6/7)^{-n} ?
The sum from n=7 to infinity of (6/7)^{-n} is diverges