What is the sum from n=1 to infinity of 4^{1/n} ?

The sum from n=1 to infinity of 4^{1/n} is diverges

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sum from n=1 to infinity of 4^{1/n}

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty n (- \frac{5}{9})^{n}

n = 0 \sum \infty \frac{1}{n ^{1.5}}

n = 1 \sum \infty \frac{9}{n ^{2} + 49}

n = 0 \sum \infty 3 (\frac{4}{5})^{n}

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

What is the sum from n=1 to infinity of 4^{1/n} ?
The sum from n=1 to infinity of 4^{1/n} is diverges