What is the sum from n=0 to infinity of 6(1/7)^{n-1} ?

The sum from n=0 to infinity of 6(1/7)^{n-1} is converges

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sum from n=0 to infinity of 6(1/7)^{n-1}

n = 0 \sum \infty 6 (\frac{1}{7})^{n - 1}

converges

Solution steps

n = 0 \sum \infty 6 (\frac{1}{7})^{n - 1}

Apply the constant multiplication rule:

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

= 6 \cdot n = 0 \sum \infty (\frac{1}{7})^{n - 1}

Apply Series Ratio Test:

converges

= 6 converges

= converges

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

n = 2 \sum \infty n^{3}

n = 0 \sum \infty \frac{n}{10 n ^{2} + 1}

n = 0 \sum \infty \frac{1}{5 ^{n} - n}

n = 1 \sum \infty 7 (0.1)^{n - 1}

What is the sum from n=0 to infinity of 6(1/7)^{n-1} ?
The sum from n=0 to infinity of 6(1/7)^{n-1} is converges