What is the sum from n=1 to infinity of 5/(3^{2n)} ?

The sum from n=1 to infinity of 5/(3^{2n)} is converges

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sum from n=1 to infinity of 5/(3^{2n)}

n = 1 \sum \infty \frac{5}{3 ^{2 n}}

converges

Solution steps

n = 1 \sum \infty \frac{5}{3 ^{2 n}}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{1}{3 ^{2 n}}

Apply Series Ratio Test:

converges

= 5 converges

= converges

y = 3 x^{3} - x^{2} + 6 x + 4

\int 2 x + 1 d x

(8 x^{3} + 2 x^{2})^{4}

\int_{- 4}^{4} 16 - x^{2} d x

\frac{d y}{d x} = 5 y

What is the sum from n=1 to infinity of 5/(3^{2n)} ?
The sum from n=1 to infinity of 5/(3^{2n)} is converges