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

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

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

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

\frac{3}{4}

Solution steps

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}

k = 1 \sum \infty 2^{- k}

k = 5 \sum \infty (\frac{e}{π})^{k - 1}

k = 4 \sum \infty \frac{1}{( k - 1 ) ^{4}}

n = 0 \sum \infty 6 (\frac{2}{3})^{n}

n = 0 \sum \infty 2 (\frac{1}{2})^{n}

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