What is the sum from n=1 to infinity of 4(0.8)^{n-1} ?

The sum from n=1 to infinity of 4(0.8)^{n-1} is converges

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sum from n=1 to infinity of 4(0.8)^{n-1}

n = 1 \sum \infty 4 (0.8)^{n - 1}

converges

Solution steps

n = 1 \sum \infty 4 \cdot 0. 8^{n - 1}

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \sum \infty 0. 8^{n - 1}

Apply Series Ratio Test:

converges

= 4 converges

= converges

d x + e^{3 x} d y = 0

\frac{d ^{2}}{d x ^{2}} (- \frac{5}{x})

\frac{d}{d x} (\frac{- 10}{x ^{3}})

e^{x} y d y - (e^{- y} + e^{2 x - y}) d x = 0

\frac{d}{d x} (lo g_{e} (x^{2} + 1))

What is the sum from n=1 to infinity of 4(0.8)^{n-1} ?
The sum from n=1 to infinity of 4(0.8)^{n-1} is converges