What is the sum from n=0 to infinity of 4/(7^{n-1)} ?

The sum from n=0 to infinity of 4/(7^{n-1)} is converges

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sum from n=0 to infinity of 4/(7^{n-1)}

n = 0 \sum \infty \frac{4}{7 ^{n - 1}}

converges

Solution steps

n = 0 \sum \infty \frac{4}{7 ^{n - 1}}

Apply the constant multiplication rule:

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

= 4 \cdot n = 0 \sum \infty \frac{1}{7 ^{n - 1}}

Apply Series Ratio Test:

converges

= 4 converges

= converges

\frac{\partial}{\partial x} (- sin (x y))

y = 0.5 x - 3

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

\int \frac{3 x ^{2}}{( 2 x ^{3} + 1 ) ^{4}} d x

\frac{d y}{d x} = x x - 6

What is the sum from n=0 to infinity of 4/(7^{n-1)} ?
The sum from n=0 to infinity of 4/(7^{n-1)} is converges