What is the sum from n=2 to infinity of 8/(nln(n)) ?

The sum from n=2 to infinity of 8/(nln(n)) is diverges

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sum from n=2 to infinity of 8/(nln(n))

n = 2 \sum \infty \frac{8}{n ln ( n )}

diverges

Solution steps

n = 2 \sum \infty \frac{8}{n ln ( n )}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 8 \cdot n = 2 \sum \infty \frac{1}{n ln ( n )}

Apply Series Integral Test:

diverges

= 8 diverges

= diverges

\frac{d y}{d x} = \frac{e ^{2 x - y}}{e ^{x + y}}

\int \frac{1}{1 - 2 x} - cos (\frac{x}{3}) d x

\frac{\partial}{\partial x} (yz e^{x})

y^{^{''}} - y^{^{'}} - 2 y = - 4 t + 6 t^{2}

What is the sum from n=2 to infinity of 8/(nln(n)) ?
The sum from n=2 to infinity of 8/(nln(n)) is diverges