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

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

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

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

converges

Solution steps

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

Simplify

(\frac{1}{2})^{2 n - 1} : (\frac{1}{2})^{2 n} (\frac{1}{2})^{- 1}

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

(\frac{1}{2})^{- 1} = 2

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

Apply Series Ratio Test:

converges

= 2 converges

= converges

\frac{\partial}{\partial x} (x z)

2 x^{3} + 3 x^{2} - 12 x + 3

f (x) = (x^{2} - 4)^{2}

3 x e^{- k x}

\frac{d ^{2}}{d x ^{2}} (\frac{x ^{2}}{( 1 + x )})

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