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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Integral Test:

diverges

= 2 diverges

= diverges

n = 1 \sum \infty 5^{- 3 n}

n = 1 \sum \infty \frac{1}{e ^{4 n}}

n = 2 \sum \infty (- 0.55)^{n}

n = 1 \sum \infty \frac{2}{n ^{3} + 4}

n = 1 \sum \infty \frac{2 n}{3 n + 6}

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