What is the sum from n=0 to infinity of 8^{1/n} ?

The sum from n=0 to infinity of 8^{1/n} is diverges

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sum from n=0 to infinity of 8^{1/n}

n = 0 \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{4}{e ^{n}}

n = 1 \sum \infty \frac{1}{( 7 n + 4 ) ^{2}}

n = 0 \sum \infty n^{5}

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

n = 0 \sum \infty \frac{n ^{4}}{n + 2}

What is the sum from n=0 to infinity of 8^{1/n} ?
The sum from n=0 to infinity of 8^{1/n} is diverges