What is the sum from n=0 to infinity of 5/(3^{n-1)} ?

The sum from n=0 to infinity of 5/(3^{n-1)} is converges

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sum from n=0 to infinity of 5/(3^{n-1)}

n = 0 \sum \infty \frac{5}{3 ^{n - 1}}

converges

Solution steps

n = 0 \sum \infty \frac{5}{3 ^{n - 1}}

Apply the constant multiplication rule:

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

= 5 \cdot n = 0 \sum \infty \frac{1}{3 ^{n - 1}}

Apply Series Ratio Test:

converges

= 5 converges

= converges

x^{\frac{1}{5}}, x^{\frac{1}{2}}

x^{^{'}} + x = 0, x (0) = 1

n = 2 \sum \infty \frac{1}{n \cdot ln ( n )}

\int_{0}^{\frac{π}{2}} (sin^{2} (x)) d x

x \to 2 - lim (1)

What is the sum from n=0 to infinity of 5/(3^{n-1)} ?
The sum from n=0 to infinity of 5/(3^{n-1)} is converges