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

The sum from n=1 to infinity of 4*(4/7)^n is 16/3

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sum from n=1 to infinity of 4*(4/7)^n

n = 1 \sum \infty 4 \cdot (\frac{4}{7})^{n}

\frac{16}{3}

Solution steps

n = 1 \sum \infty 4 (\frac{4}{7})^{n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 4 \cdot n = 1 \sum \infty (\frac{4}{7})^{n}

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

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

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

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

Simplify

4 (\frac{7}{3} - (\frac{4}{7})^{0}) : \frac{16}{3}

= \frac{16}{3}

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What is the sum from n=1 to infinity of 4*(4/7)^n ?
The sum from n=1 to infinity of 4*(4/7)^n is 16/3