What is the sum from n=1 to infinity of 4/(10^n) ?

The sum from n=1 to infinity of 4/(10^n) is 4/9

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sum from n=1 to infinity of 4/(10^n)

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

\frac{4}{9}

Solution steps

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

Apply the constant multiplication rule:

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

= 4 \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}

= 4 (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}

= 4 (\frac{10}{9} - \frac{1}{1 0 ^{0}})

Simplify

4 (\frac{10}{9} - \frac{1}{1 0 ^{0}}) : \frac{4}{9}

= \frac{4}{9}

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

y = 4 - x^{2}, 2 x + y = 4

f (x) = x^{2} sin (x)

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

x \to 3 + lim (\frac{2 x}{x - 3})

What is the sum from n=1 to infinity of 4/(10^n) ?
The sum from n=1 to infinity of 4/(10^n) is 4/9