What is the sum from n=1 to infinity of 3(9/7)^{n-1} ?

The sum from n=1 to infinity of 3(9/7)^{n-1} is diverges

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sum from n=1 to infinity of 3(9/7)^{n-1}

n = 1 \sum \infty 3 (\frac{9}{7})^{n - 1}

diverges

Solution steps

n = 1 \sum \infty 3 (\frac{9}{7})^{n - 1}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty (\frac{9}{7})^{n - 1}

Apply Series Ratio Test:

diverges

= 3 diverges

= diverges

n = 1 \sum \infty \frac{cos ( πn )}{n}

n = 1 \sum \infty \frac{1}{7 n}

n = 1 \sum \infty (\frac{5}{4})^{n - 1}

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

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

What is the sum from n=1 to infinity of 3(9/7)^{n-1} ?
The sum from n=1 to infinity of 3(9/7)^{n-1} is diverges