What is the sum from n=1 to infinity of 4^{-2n+1} ?

The sum from n=1 to infinity of 4^{-2n+1} is converges

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sum from n=1 to infinity of 4^{-2n+1}

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

converges

Solution steps

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

Simplify

4^{- 2 n + 1} : 4^{- 2 n} \cdot 4

= n = 1 \sum \infty 4^{- 2 n} \cdot 4

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \sum \infty 4^{- 2 n}

Apply Series Ratio Test:

converges

= 4 converges

= converges

\frac{d}{d x} (\frac{1}{4 - x ^{2}})

x \to 1 lim ((1 - x^{2})^{1 - x})

\frac{d}{d v} (u + v)

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

\int \frac{x}{( 1 - x ^{2} )} d x

What is the sum from n=1 to infinity of 4^{-2n+1} ?
The sum from n=1 to infinity of 4^{-2n+1} is converges