What is the sum from n=0 to infinity of 9^{1/n} ?

The sum from n=0 to infinity of 9^{1/n} is diverges

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sum from n=0 to infinity of 9^{1/n}

n = 0 \sum \infty 9^{\frac{1}{n}}

diverges

Solution steps

n = 1 \sum \infty 9^{\frac{1}{n}}

Apply Series Divergence Test:

diverges

= diverges

n = 0 \sum \infty 3 (\frac{2}{5})^{n + 1}

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

n = 0 \sum \infty \frac{1}{5 + 6 n}

n = 1 \sum \infty \frac{1}{9 n - 1}

n = 1 \sum \infty 35 (- 1)^{n}

What is the sum from n=0 to infinity of 9^{1/n} ?
The sum from n=0 to infinity of 9^{1/n} is diverges