What is the sum from n=0 to infinity of 2/(3^{n+1)} ?

The sum from n=0 to infinity of 2/(3^{n+1)} is converges

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sum from n=0 to infinity of 2/(3^{n+1)}

n = 0 \sum \infty \frac{2}{3 ^{n + 1}}

converges

Solution steps

n = 0 \sum \infty \frac{2}{3 ^{n + 1}}

Apply the constant multiplication rule:

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

= 2 \cdot n = 0 \sum \infty \frac{1}{3 ^{n + 1}}

Apply Series Ratio Test:

converges

= 2 converges

= converges

y^{^{''}} - b^{2} y = 0

\int \frac{ln ( 2 ) x ^{5}}{x ^{2}} d x

\int_{0}^{4} x - (x - 2) d x

x^{^{''}} + x = 0, x (\frac{π}{2}) = 0, x^{^{'}} (\frac{π}{2}) = 1

\int a^{x} x d x

What is the sum from n=0 to infinity of 2/(3^{n+1)} ?
The sum from n=0 to infinity of 2/(3^{n+1)} is converges