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

Popular >

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

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

\frac{2}{9}

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}

\int (\frac{x + 6}{x}) d x

\frac{d}{d x} (\frac{2 x + x ^{4}}{( 1 - x ^{3} ) ^{2}})

\int ln (sin (x)) d x

\frac{\partial}{\partial y} (2 e^{- l n (y)} y)

\frac{d ^{2}}{d x ^{2}} (cos (- 1.5 x))

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