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

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

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sum from k=1 to infinity of k^{1/k}

k = 1 \sum \infty k^{\frac{1}{k}}

diverges

Solution steps

k = 1 \sum \infty k^{\frac{1}{k}}

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty \frac{1}{n ^{3} + 5}

n = 1 \sum \infty \frac{1}{n ^{3} - 5 n ^{2}}

k = 1 \sum \infty \frac{k ^{2}}{4 ^{k}}

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

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

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