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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 3 converges

= converges

x \to 1 - lim (1 - x^{2})

\frac{d}{d x} (\frac{x ^{2} - 6 x + 7}{x ^{2} + 9})

\frac{d}{d x} (\frac{x ^{3}}{1 - x})

f (x) = x^{2} + 2, - 3, 1

\frac{d y}{d x}, y = e^{9 x^{5}}

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