What is the sum from n=1 to infinity of 10(-3/8)^n ?

The sum from n=1 to infinity of 10(-3/8)^n is converges

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sum from n=1 to infinity of 10(-3/8)^n

n = 1 \sum \infty 10 (- \frac{3}{8})^{n}

converges

Solution steps

n = 1 \sum \infty 10 (- \frac{3}{8})^{n}

Apply the constant multiplication rule:

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

= 10 \cdot n = 1 \sum \infty (- \frac{3}{8})^{n}

Apply Series Ratio Test:

converges

= 10 converges

= converges

n = 1 \sum \infty \frac{6 ^{n}}{n ^{6}}

n = 0 \sum \infty (1 + \frac{9}{n})^{n}

k = 1 \sum \infty \frac{k ^{4}}{k !}

n = 0 \sum \infty \frac{3}{2 n + 2}

n = 1 \sum \infty n^{- 5}

What is the sum from n=1 to infinity of 10(-3/8)^n ?
The sum from n=1 to infinity of 10(-3/8)^n is converges