What is the sum from n=1 to infinity of 2/(10^n) ?

The sum from n=1 to infinity of 2/(10^n) is 2/9

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sum from n=1 to infinity of 2/(10^n)

n = 1 \sum \infty \frac{2}{1 0 ^{n}}

\frac{2}{9}

Decimal

0.22222 \dots

Solution steps

n = 1 \sum \infty \frac{2}{1 0 ^{n}}

Apply the constant multiplication rule:

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

= 2 \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}

= 2 (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}

= 2 (\frac{10}{9} - \frac{1}{1 0 ^{0}})

Simplify

2 (\frac{10}{9} - \frac{1}{1 0 ^{0}}) : \frac{2}{9}

= \frac{2}{9}

What is the sum from n=1 to infinity of 2/(10^n) ?
The sum from n=1 to infinity of 2/(10^n) is 2/9