What is the sum from n=1 to infinity of \sqrt[n]{n} ?

The sum from n=1 to infinity of \sqrt[n]{n} is diverges

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sum from n=1 to infinity of \sqrt[n]{n}

n = 1 \sum \infty n n

diverges

Solution steps

n = 1 \sum \infty n n

Apply Series Divergence Test:

diverges

= diverges

f (x) = tan (\frac{3 π}{4} x)

\int a^{- x} d x

x \to 3 - lim (arcsin (\frac{x}{3}))

\frac{1}{3 x}

\int 15 sin^{2} (x) d x

What is the sum from n=1 to infinity of \sqrt[n]{n} ?
The sum from n=1 to infinity of \sqrt[n]{n} is diverges