What is the sum from n=5 to infinity of (4/5)^{-n} ?

The sum from n=5 to infinity of (4/5)^{-n} is diverges

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sum from n=5 to infinity of (4/5)^{-n}

n = 5 \sum \infty (\frac{4}{5})^{- n}

diverges

Solution steps

n = 5 \sum \infty (\frac{4}{5})^{- n}

Apply Series Ratio Test:

diverges

= diverges

9 t

f (x) = (x - 1)^{2} + 2, x = - 1, x = 2

\int \frac{5 x ^{2} - 12 x - 12}{x ^{3} - 4 x} d x

\int \frac{2 x ^{5} - x ^{3} - 1}{x ^{3} - 4 x} d x

x \to 1 + lim (\frac{2 x}{x - 1})

What is the sum from n=5 to infinity of (4/5)^{-n} ?
The sum from n=5 to infinity of (4/5)^{-n} is diverges