What is the sum from n=1 to infinity of (-4/5)^{n-1} ?

The sum from n=1 to infinity of (-4/5)^{n-1} is converges

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sum from n=1 to infinity of (-4/5)^{n-1}

n = 1 \sum \infty (- \frac{4}{5})^{n - 1}

converges

Solution steps

n = 1 \sum \infty (- \frac{4}{5})^{n - 1}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty (\frac{3}{4})^{- n}

n = 0 \sum \infty \frac{1}{ln ( 1 + n )}

n = 1 \sum \infty \frac{1}{4 + 3 ^{n}}

k = 1 \sum \infty \frac{1}{2 ^{k + 1}}

n = 0 \sum \infty \frac{4}{9 ^{n}}

What is the sum from n=1 to infinity of (-4/5)^{n-1} ?
The sum from n=1 to infinity of (-4/5)^{n-1} is converges