What is the sum from n=1 to infinity of 2(0.1)^n ?

The sum from n=1 to infinity of 2(0.1)^n is 0.22222…

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sum from n=1 to infinity of 2(0.1)^n

n = 1 \sum \infty 2 (0.1)^{n}

0.22222 \dots

Solution steps

n = 1 \sum \infty 2 \cdot 0. 1^{n}

Apply the constant multiplication rule:

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

= 2 \cdot 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}

= 2 (n = 0 \sum \infty 0. 1^{n} - 0. 1^{0})

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

= 2 (1.11111 \dots - 0. 1^{0})

Simplify

= 0.22222 \dots

\frac{d}{d x} \frac{1}{f ( x ) x ^{^{'}} ( f ( x ) ( x ) ^{- 1} ( x ) )}

\frac{d}{d x} (\frac{cos ( x )}{2})

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{2 t}

x \to 4 lim (\frac{x}{x ^{2} - 16})

x \to 1 lim (\frac{x ^{2} - 3}{x - 6})

What is the sum from n=1 to infinity of 2(0.1)^n ?
The sum from n=1 to infinity of 2(0.1)^n is 0.22222…