What is the sum from n=1 to infinity of 1/((0.8)^n) ?

The sum from n=1 to infinity of 1/((0.8)^n) is diverges

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sum from n=1 to infinity of 1/((0.8)^n)

n = 1 \sum \infty \frac{1}{( 0.8 ) ^{n}}

diverges

Solution steps

n = 1 \sum \infty \frac{1}{0. 8 ^{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}{0. 8 ^{n}} - \frac{1}{0. 8 ^{0}}

n = 0 \sum \infty \frac{1}{0. 8 ^{n}} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{0}^{2} x^{2} 4 - x^{2} d x

f (x) = \frac{- 8 x}{x ^{2} + 1}

3 t + 2

x \to - 2 lim (x^{2} + 2 x - 3)

4 cos (x)

What is the sum from n=1 to infinity of 1/((0.8)^n) ?
The sum from n=1 to infinity of 1/((0.8)^n) is diverges