What is the sum from n=0 to infinity of [7(0.1)^n] ?

The sum from n=0 to infinity of [7(0.1)^n] is 7.77777…

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sum from n=0 to infinity of [7(0.1)^n]

n = 0 \sum \infty [7 (0.1)^{n}]

7.77777 \dots

Solution steps

n = 0 \sum \infty 7 \cdot 0. 1^{n}

Apply the constant multiplication rule:

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

= 7 \cdot n = 0 \sum \infty 0. 1^{n}

Apply Series Geometric Test:

1.11111 \dots

= 7 \cdot 1.11111 \dots

Simplify

= 7.77777 \dots

n = 1 \sum \infty \frac{3}{n} - \frac{3}{n + 2}

n = 2 \sum \infty n e^{- 3 n^{2}}

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

k = 3 \sum \infty \frac{1}{4 k ^{2} - 1}

k = 1 \sum \infty \frac{1}{k} - \frac{1}{k + 1}

What is the sum from n=0 to infinity of [7(0.1)^n] ?
The sum from n=0 to infinity of [7(0.1)^n] is 7.77777…