What is the sum from n=1 to infinity of (-0.01+1)^n ?

The sum from n=1 to infinity of (-0.01+1)^n is 99

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sum from n=1 to infinity of (-0.01+1)^n

n = 1 \sum \infty (- 0.01 + 1)^{n}

99

Solution steps

n = 1 \sum \infty (- 0.01 + 1)^{n}

Simplify

= n = 1 \sum \infty 0.9 9^{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 0.9 9^{n} - 0.9 9^{0}

n = 0 \sum \infty 0.9 9^{n} = 100

= 100 - 0.9 9^{0}

Simplify

= 99

What is the sum from n=1 to infinity of (-0.01+1)^n ?
The sum from n=1 to infinity of (-0.01+1)^n is 99