What is the sum from n=1 to infinity of (3^n)/(7^n) ?

The sum from n=1 to infinity of (3^n)/(7^n) is 3/4

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sum from n=1 to infinity of (3^n)/(7^n)

n = 1 \sum \infty \frac{3 ^{n}}{7 ^{n}}

\frac{3}{4}

Solution steps

n = 1 \sum \infty \frac{3 ^{n}}{7 ^{n}}

Apply exponent rule:

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

= n = 1 \sum \infty (\frac{3}{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{3}{7})^{n} - (\frac{3}{7})^{0}

n = 0 \sum \infty (\frac{3}{7})^{n} = \frac{7}{4}

= \frac{7}{4} - (\frac{3}{7})^{0}

Simplify

\frac{7}{4} - (\frac{3}{7})^{0} : \frac{3}{4}

= \frac{3}{4}

n = 3 \sum \infty \frac{6}{n ^{3}}

n = 1 \sum \infty 4 (1.3)^{n}

n = 0 \sum \infty n^{^{''}} - n^{^{'}}

n = 0 \sum \infty (\frac{99}{100})^{n}

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

What is the sum from n=1 to infinity of (3^n)/(7^n) ?
The sum from n=1 to infinity of (3^n)/(7^n) is 3/4