What is the sum from n=1 to infinity of 5/(10^n) ?

The sum from n=1 to infinity of 5/(10^n) is 5/9

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sum from n=1 to infinity of 5/(10^n)

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

\frac{5}{9}

Solution steps

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

Apply the constant multiplication rule:

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

= 5 \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}

= 5 (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}

= 5 (\frac{10}{9} - \frac{1}{1 0 ^{0}})

Simplify

5 (\frac{10}{9} - \frac{1}{1 0 ^{0}}) : \frac{5}{9}

= \frac{5}{9}

y = 14 x

\int 12 tan^{11} (x) sec^{2} (x) d x

\int_{1}^{5} \frac{x ^{2} + 2}{6 x - x ^{2}} d x

\int 3 x^{2} + 4 x + 1 d x

3 \frac{d y}{d x} + 15 y = 5

What is the sum from n=1 to infinity of 5/(10^n) ?
The sum from n=1 to infinity of 5/(10^n) is 5/9