What is the sum from n=1 to infinity of 8/(10^n) ?

The sum from n=1 to infinity of 8/(10^n) is 8/9

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sum from n=1 to infinity of 8/(10^n)

n = 1 \sum \infty \frac{8}{1 0 ^{n}}

\frac{8}{9}

Solution steps

n = 1 \sum \infty \frac{8}{1 0 ^{n}}

Apply the constant multiplication rule:

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

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

n = 0 \sum \infty \frac{1}{1 0 ^{n}} = \frac{10}{9}

= 8 (\frac{10}{9} - \frac{1}{1 0 ^{0}})

Simplify

8 (\frac{10}{9} - \frac{1}{1 0 ^{0}}) : \frac{8}{9}

= \frac{8}{9}

n = 1 \sum \infty cos (\frac{15}{n})

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

n = 1 \sum \infty 2^{n + 1}

n = 0 \sum \infty n \cdot 2^{n}

n = 1 \sum \infty (- π)^{n}

What is the sum from n=1 to infinity of 8/(10^n) ?
The sum from n=1 to infinity of 8/(10^n) is 8/9