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

The sum from i=1 to infinity of (1/2)^{i-1} is converges

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

y^{^{''}} - 3 y^{^{'}} = 2 e^{x} sin (2 x)

f (x) = \frac{x - 7}{9 x ^{2} e ^{x}}

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

\frac{d}{d x} ((3 x + 2) (4 x - 3))

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