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

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

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

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

6

Solution steps

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

Apply exponent rule:

\frac{a ^{c}}{b ^{c}} = (\frac{a}{b})^{c}

= 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

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