What is the sum from n=1 to infinity of (1/2)^{99-1} ?

The sum from n=1 to infinity of (1/2)^{99-1} is diverges

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sum from n=1 to infinity of (1/2)^{99-1}

n = 1 \sum \infty (\frac{1}{2})^{99 - 1}

diverges

Solution steps

n = 1 \sum \infty (\frac{1}{2})^{99 - 1}

Every infinite sum of a non-zero constant diverges

= diverges

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\frac{\partial}{\partial y} (e^{- 3 x} cos (3 y))

f (x) = ln (1 - x)

y^{^{'}} = - k y^{\frac{1}{2}}

What is the sum from n=1 to infinity of (1/2)^{99-1} ?
The sum from n=1 to infinity of (1/2)^{99-1} is diverges