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

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

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

n = 1 \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

\frac{d}{d x} (\frac{4 x ^{2} + 4 x + 2}{x})

\int_{1}^{3} 2 x e^{- 4 x} d x

\int 3 x^{2} + 4 x + 1 d x

\frac{d y}{d x} + \frac{1}{x + 4} y = x^{- 2}

f (x) = \frac{9 x}{e ^{x}}

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