What is the sum from n=1 to infinity of 2/(3^{2n)} ?

The sum from n=1 to infinity of 2/(3^{2n)} is converges

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sum from n=1 to infinity of 2/(3^{2n)}

n = 1 \sum \infty \frac{2}{3 ^{2 n}}

converges

Solution steps

n = 1 \sum \infty \frac{2}{3 ^{2 n}}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty \frac{1}{3 ^{2 n}}

Apply Series Ratio Test:

converges

= 2 converges

= converges

\frac{d}{d x} (\frac{11}{x})

\int \frac{x ^{2}}{x ^{2} + 3} d x

y^{^{'''}} + 4 y^{^{'}} = sec (2 x)

\frac{d y}{d x} = 3 y (5 - y)

What is the sum from n=1 to infinity of 2/(3^{2n)} ?
The sum from n=1 to infinity of 2/(3^{2n)} is converges