What is the sum from n=1 to infinity of 6/(5^{n-1)} ?

The sum from n=1 to infinity of 6/(5^{n-1)} is converges

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sum from n=1 to infinity of 6/(5^{n-1)}

n = 1 \sum \infty \frac{6}{5 ^{n - 1}}

converges

Solution steps

n = 1 \sum \infty \frac{6}{5 ^{n - 1}}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty \frac{1}{5 ^{n - 1}}

Apply Series Ratio Test:

converges

= 6 converges

= converges

\frac{d}{d x} (4 (x^{2} + y^{2}))

x \to 2 lim (3 x)

y = (2 x + 1)^{3} (3 x + 4)^{- 5}

\int (x - 1)^{2} d x

x^{2} d y = (x^{2} + x y + y^{2}) d x

What is the sum from n=1 to infinity of 6/(5^{n-1)} ?
The sum from n=1 to infinity of 6/(5^{n-1)} is converges