What is the sum from n=1 to infinity of 7/(10^n) ?

The sum from n=1 to infinity of 7/(10^n) is 7/9

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sum from n=1 to infinity of 7/(10^n)

n = 1 \sum \infty \frac{7}{1 0 ^{n}}

\frac{7}{9}

Solution steps

n = 1 \sum \infty \frac{7}{1 0 ^{n}}

Apply the constant multiplication rule:

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

= 7 \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}

= 7 (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}

= 7 (\frac{10}{9} - \frac{1}{1 0 ^{0}})

Simplify

7 (\frac{10}{9} - \frac{1}{1 0 ^{0}}) : \frac{7}{9}

= \frac{7}{9}

f (x) = 2 x^{3} + 36 x^{2} + 192 x + 8

x \to 0 + lim (\frac{e ^{x}}{1 - e ^{- x}})

\frac{d}{d x} (4 e^{6 x})

f (x) = sin (3 x + 1)

\frac{d}{d x} ((4 x)^{6 x})

What is the sum from n=1 to infinity of 7/(10^n) ?
The sum from n=1 to infinity of 7/(10^n) is 7/9