What is the sum from n=0 to infinity of 4(2/3)^{n-1} ?

The sum from n=0 to infinity of 4(2/3)^{n-1} is converges

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sum from n=0 to infinity of 4(2/3)^{n-1}

n = 0 \sum \infty 4 (\frac{2}{3})^{n - 1}

converges

Solution steps

n = 0 \sum \infty 4 (\frac{2}{3})^{n - 1}

Apply the constant multiplication rule:

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

= 4 \cdot n = 0 \sum \infty (\frac{2}{3})^{n - 1}

Apply Series Ratio Test:

converges

= 4 converges

= converges

\int \frac{16 x ^{2} - 1}{x} d x

\frac{d}{d x} (x^{2} (x - 12)^{2})

\frac{ln ( x )}{1 + ln ( x )}

\int 4^{x} d x

f (x) = 5 x^{2} + 4 x - 10

What is the sum from n=0 to infinity of 4(2/3)^{n-1} ?
The sum from n=0 to infinity of 4(2/3)^{n-1} is converges