What is the sum from n=1 to infinity of (-9/8)^{n-1} ?

The sum from n=1 to infinity of (-9/8)^{n-1} is diverges

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sum from n=1 to infinity of (-9/8)^{n-1}

n = 1 \sum \infty (- \frac{9}{8})^{n - 1}

diverges

Solution steps

n = 1 \sum \infty (- \frac{9}{8})^{n - 1}

Apply Series Ratio Test:

diverges

= diverges

n = 0 \sum \infty \frac{1}{1 + 3 n ^{4}}

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

n = 0 \sum \infty 2 \cdot 2^{n}

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

n = 0 \sum \infty \frac{11}{2 n - 1}

What is the sum from n=1 to infinity of (-9/8)^{n-1} ?
The sum from n=1 to infinity of (-9/8)^{n-1} is diverges