What is the sum from n=2 to infinity of 1/(2nln(n)) ?

The sum from n=2 to infinity of 1/(2nln(n)) is diverges

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sum from n=2 to infinity of 1/(2nln(n))

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Integral Test:

diverges

= \frac{1}{2} diverges

= diverges

ln (x^{\frac{1}{2}})

\frac{d}{d x} (x (a cos (x) + b sin (x)))

\int_{- 1}^{e - 2} \frac{t}{t ^{2} + 2} d t

\frac{\partial}{\partial x} (x cos (x) y)

\frac{d}{d x} (x co x)

What is the sum from n=2 to infinity of 1/(2nln(n)) ?
The sum from n=2 to infinity of 1/(2nln(n)) is diverges