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

The sum from n=1 to infinity of (6/7)^n is 6

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sum from n=1 to infinity of (6/7)^n

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

6

Solution steps

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

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

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

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

= 7 - (\frac{6}{7})^{0}

Simplify

= 6

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What is the sum from n=1 to infinity of (6/7)^n ?
The sum from n=1 to infinity of (6/7)^n is 6