What is the sum from n=1 to infinity of 1/(1.05^n)*1 ?

The sum from n=1 to infinity of 1/(1.05^n)*1 is 20

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sum from n=1 to infinity of 1/(1.05^n)*1

n = 1 \sum \infty \frac{1}{1.0 5 ^{n}} \cdot 1

20

Solution steps

n = 1 \sum \infty \frac{1}{1.0 5 ^{n}} \cdot 1

Simplify

= n = 1 \sum \infty \frac{1}{1.0 5 ^{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}{1.0 5 ^{n}} - \frac{1}{1.0 5 ^{0}}

n = 0 \sum \infty \frac{1}{1.0 5 ^{n}} = 21

= 21 - \frac{1}{1.0 5 ^{0}}

Simplify

= 20

What is the sum from n=1 to infinity of 1/(1.05^n)*1 ?
The sum from n=1 to infinity of 1/(1.05^n)*1 is 20