What is the sum from n=1 to infinity of 6(1/8)^{n-1} ?

The sum from n=1 to infinity of 6(1/8)^{n-1} is converges

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sum from n=1 to infinity of 6(1/8)^{n-1}

n = 1 \sum \infty 6 (\frac{1}{8})^{n - 1}

converges

Solution steps

n = 1 \sum \infty 6 (\frac{1}{8})^{n - 1}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty (\frac{1}{8})^{n - 1}

Apply Series Ratio Test:

converges

= 6 converges

= converges

n = 0 \sum \infty 3 n ln (1 + n)

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

n = 1 \sum \infty (\frac{5}{6})^{4 n - 5}

n = 2 \sum \infty (\frac{1}{3})^{3 n}

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

What is the sum from n=1 to infinity of 6(1/8)^{n-1} ?
The sum from n=1 to infinity of 6(1/8)^{n-1} is converges