What is the sum from n=0 to infinity of (-0.2)^n ?

The sum from n=0 to infinity of (-0.2)^n is 0.83333…

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sum from n=0 to infinity of (-0.2)^n

n = 0 \sum \infty (- 0.2)^{n}

0.83333 \dots

Solution steps

n = 0 \sum \infty (- 0.2)^{n}

Apply Series Geometric Test:

0.83333 \dots

= 0.83333 \dots

n = 0 \sum \infty \frac{1}{( 1 + n )}

n = 0 \sum \infty cos (\frac{3}{n ^{2}})

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

n = 1 \sum \infty \frac{4}{n ^{2} + n}

n = 1 \sum \infty \frac{4 n}{4 n + 1}

What is the sum from n=0 to infinity of (-0.2)^n ?
The sum from n=0 to infinity of (-0.2)^n is 0.83333…