What is the sum from n=1 to infinity of (4/9)^{2n+1} ?

The sum from n=1 to infinity of (4/9)^{2n+1} is converges

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sum from n=1 to infinity of (4/9)^{2n+1}

n = 1 \sum \infty (\frac{4}{9})^{2 n + 1}

converges

Solution steps

n = 1 \sum \infty (\frac{4}{9})^{2 n + 1}

Simplify

(\frac{4}{9})^{2 n + 1} : (\frac{4}{9})^{2 n} \frac{4}{9}

= n = 1 \sum \infty (\frac{4}{9})^{2 n} \frac{4}{9}

Apply the constant multiplication rule:

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

= \frac{4}{9} \cdot n = 1 \sum \infty (\frac{4}{9})^{2 n}

Apply Series Ratio Test:

converges

= \frac{4}{9} converges

= converges

f (x) = x^{2} + x, at x = 2

y^{^{''}} + 3 y^{^{'}} + 3 y = 0

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

\frac{d y}{d x} + 3 x y^{3} = 0

\int_{0}^{r} x arcsin (\frac{x}{2 r}) d x

What is the sum from n=1 to infinity of (4/9)^{2n+1} ?
The sum from n=1 to infinity of (4/9)^{2n+1} is converges