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

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

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sum from n=1 to infinity of 8^{1/n}

n = 1 \sum \infty 8^{\frac{1}{n}}

diverges

Solution steps

n = 1 \sum \infty 8^{\frac{1}{n}}

Apply Series Divergence Test:

diverges

= diverges

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

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

k = 1 \sum \infty k e^{- 3 k^{2}}

n = 0 \sum \infty \frac{1 + n}{n !}

n = 2 \sum \infty \frac{3 ^{n}}{n !}

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