What is the sum from n=1 to infinity of 5(0.8)^{n-1} ?

The sum from n=1 to infinity of 5(0.8)^{n-1} is converges

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sum from n=1 to infinity of 5(0.8)^{n-1}

n = 1 \sum \infty 5 (0.8)^{n - 1}

converges

Solution steps

n = 1 \sum \infty 5 \cdot 0. 8^{n - 1}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty 0. 8^{n - 1}

Apply Series Ratio Test:

converges

= 5 converges

= converges

y^{^{''}} + 6 y^{^{'}} = 0

\int_{2}^{\infty} y e^{- 3 y} d y

x \to - 2 - lim (- 2 x + 6)

\int_{0}^{1} \frac{55}{4 y - 1} d y

y = lo g_{3} (6 x)

What is the sum from n=1 to infinity of 5(0.8)^{n-1} ?
The sum from n=1 to infinity of 5(0.8)^{n-1} is converges