What is the sum from n=0 to infinity of (3/n)^{1/n} ?

The sum from n=0 to infinity of (3/n)^{1/n} is diverges

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sum from n=0 to infinity of (3/n)^{1/n}

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

n = 2 \sum \infty \frac{n ^{4}}{n ^{3}}

n = 2 \sum \infty n e^{- 5 n}

n = 0 \sum \infty e^{3 n}

n = 0 \sum \infty \frac{3 ^{n}}{n ^{2}}

n = 1 \sum \infty \frac{3 n}{n ^{2} + 3}

What is the sum from n=0 to infinity of (3/n)^{1/n} ?
The sum from n=0 to infinity of (3/n)^{1/n} is diverges