What is the sum from n=1 to infinity of-8(4/5)^{n-1} ?

The sum from n=1 to infinity of-8(4/5)^{n-1} is converges

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sum from n=1 to infinity of-8(4/5)^{n-1}

n = 1 \sum \infty - 8 (\frac{4}{5})^{n - 1}

converges

Solution steps

n = 1 \sum \infty - 8 (\frac{4}{5})^{n - 1}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= - 8 \cdot n = 1 \sum \infty (\frac{4}{5})^{n - 1}

Apply Series Ratio Test:

converges

= - 8 converges

= converges

n = 0 \sum \infty (\frac{- 2}{3})^{n}

n = 7 \sum \infty 2^{- n}

n = 0 \sum \infty (\frac{1}{2})^{n - 1}

n = 0 \sum \infty sin (\frac{1}{n ^{1}})

n = 0 \sum \infty (\frac{11}{12})^{n}

What is the sum from n=1 to infinity of-8(4/5)^{n-1} ?
The sum from n=1 to infinity of-8(4/5)^{n-1} is converges