What is the sum from n=1 to infinity of 90(0.1)^n ?

The sum from n=1 to infinity of 90(0.1)^n is 10

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sum from n=1 to infinity of 90(0.1)^n

n = 1 \sum \infty 90 (0.1)^{n}

10

Solution steps

n = 1 \sum \infty 90 \cdot 0. 1^{n}

Apply the constant multiplication rule:

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

= 90 \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}

= 90 (n = 0 \sum \infty 0. 1^{n} - 0. 1^{0})

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

= 90 (1.11111 \dots - 0. 1^{0})

Simplify

= 10

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

n = 1 \sum \infty \frac{1}{5 + e ^{- n}}

m = 1 \sum \infty \frac{10}{m ^{8} + 3}

n = 1 \sum \infty 2 ln (\frac{n}{n + 1})

n = 1 \sum \infty \frac{n}{n ^{2} + 6}

What is the sum from n=1 to infinity of 90(0.1)^n ?
The sum from n=1 to infinity of 90(0.1)^n is 10