What is the sum from n=0 to infinity of 8*(2/7)^n ?

The sum from n=0 to infinity of 8*(2/7)^n is 56/5

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sum from n=0 to infinity of 8*(2/7)^n

n = 0 \sum \infty 8 \cdot (\frac{2}{7})^{n}

\frac{56}{5}

Solution steps

n = 0 \sum \infty 8 (\frac{2}{7})^{n}

Apply the constant multiplication rule:

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

= 8 \cdot n = 0 \sum \infty (\frac{2}{7})^{n}

Apply Series Geometric Test:

\frac{7}{5}

= 8 \cdot \frac{7}{5}

Simplify

8 \cdot \frac{7}{5} : \frac{56}{5}

= \frac{56}{5}

k = 1 \sum \infty k e^{- 7 k^{2}}

n = 0 \sum \infty \frac{n !}{n}

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

n = 1 \sum \infty \frac{5 ^{n}}{n ^{2}}

n = 4 \sum \infty \frac{3}{n ^{2} - n}

What is the sum from n=0 to infinity of 8*(2/7)^n ?
The sum from n=0 to infinity of 8*(2/7)^n is 56/5