What is the sum from n=1 to infinity of 1/(1.1^n) ?

The sum from n=1 to infinity of 1/(1.1^n) is 10

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sum from n=1 to infinity of 1/(1.1^n)

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

10

Solution steps

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

n = 0 \sum \infty \frac{1}{1. 1 ^{n}} = 11

= 11 - \frac{1}{1. 1 ^{0}}

Simplify

= 10

k = 1 \sum \infty \frac{1}{16 k ^{2} - 4}

n = 1 \sum \infty \frac{n}{n ^{2} - 2 n + 5}

n = 1 \sum \infty \frac{7}{1 + n ^{2}}

n = 0 \sum \infty 3

n = 1 \sum \infty \frac{2 n}{7 n - 9}

What is the sum from n=1 to infinity of 1/(1.1^n) ?
The sum from n=1 to infinity of 1/(1.1^n) is 10