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

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

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

n = 1 \sum \infty (\frac{3}{4})^{n}

3

Solution steps

n = 1 \sum \infty (\frac{3}{4})^{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}{4})^{n} - (\frac{3}{4})^{0}

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

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

Simplify

= 3

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