What is the sum from n=2 to infinity of (ln^2(n))/n ?

The sum from n=2 to infinity of (ln^2(n))/n is diverges

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sum from n=2 to infinity of (ln^2(n))/n

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

\int_{0}^{x} \frac{2}{1 + t} d t

\frac{d}{d x} (cos^{3} (5 x + 1))

y^{^{''}} + 4 y = - 6, y (\frac{π}{8}) = - 12, y^{^{'}} (\frac{π}{8}) = 2

\frac{d}{d x} ((x - 1) e^{- x})

\frac{d}{d x} (e^{3 x y})

What is the sum from n=2 to infinity of (ln^2(n))/n ?
The sum from n=2 to infinity of (ln^2(n))/n is diverges