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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

\frac{d}{d x} (\frac{16}{x})

y^{^{''}} - 3 y^{^{'}} = - 9

\int \frac{x ^{3}}{25 - x ^{2}} d x

\int_{0}^{2} 3 y 4 - y^{2} d y

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = 3

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