What is the sum from n=1 to infinity of (0.1)^n ?

The sum from n=1 to infinity of (0.1)^n is 0.11111…

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sum from n=1 to infinity of (0.1)^n

n = 1 \sum \infty (0.1)^{n}

0.11111 \dots

Solution steps

n = 1 \sum \infty 0. 1^{n}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= n = 0 \sum \infty 0. 1^{n} - 0. 1^{0}

n = 0 \sum \infty 0. 1^{n} = 1.11111 \dots

= 1.11111 \dots - 0. 1^{0}

Simplify

= 0.11111 \dots

\frac{d}{d x} (\frac{2 x + 5}{x ^{2} - 3})

f (x) = - 3 x + 4

(0, 0), (3, 4)

- \frac{3}{4 s ^{4}} + \frac{10}{3 s ^{3}}

x \to - 1 lim (\frac{( x - 2 )}{x + 1})

What is the sum from n=1 to infinity of (0.1)^n ?
The sum from n=1 to infinity of (0.1)^n is 0.11111…