What is the sum from n=1 to infinity of 8*(1/5)^n ?

The sum from n=1 to infinity of 8*(1/5)^n is 2

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sum from n=1 to infinity of 8*(1/5)^n

n = 1 \sum \infty 8 \cdot (\frac{1}{5})^{n}

2

Solution steps

n = 1 \sum \infty 8 (\frac{1}{5})^{n}

Apply the constant multiplication rule:

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

= 8 \cdot n = 1 \sum \infty (\frac{1}{5})^{n}

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

= 8 (n = 0 \sum \infty (\frac{1}{5})^{n} - (\frac{1}{5})^{0})

n = 0 \sum \infty (\frac{1}{5})^{n} = \frac{5}{4}

= 8 (\frac{5}{4} - (\frac{1}{5})^{0})

Simplify

8 (\frac{5}{4} - (\frac{1}{5})^{0}) : 2

= 2

\int \frac{7 x + 1}{x ( x ^{2} - 1 )} d x

\int_{0}^{3} \frac{1}{1 + x} d x

\int \frac{z ^{2}}{1 ^{3}} + z^{3}

\int 20 x^{3} - 12 x^{2} + 6 x d x

n = 1 \sum \infty \frac{7}{n ^{2} + 9}

What is the sum from n=1 to infinity of 8*(1/5)^n ?
The sum from n=1 to infinity of 8*(1/5)^n is 2