What is the sum from n=2 to infinity of 4/(nln(n)) ?

The sum from n=2 to infinity of 4/(nln(n)) is diverges

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sum from n=2 to infinity of 4/(nln(n))

n = 2 \sum \infty \frac{4}{n ln ( n )}

diverges

Solution steps

n = 2 \sum \infty \frac{4}{n ln ( n )}

Apply the constant multiplication rule:

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

= 4 \cdot n = 2 \sum \infty \frac{1}{n ln ( n )}

Apply Series Integral Test:

diverges

= 4 diverges

= diverges

n = 1 \sum \infty \frac{6 ^{n}}{n}

n = 0 \sum \infty \frac{1}{( n ( n + 1 ) )}

n = 0 \sum \infty 1 + (- 1)^{n}

n = 0 \sum \infty ln (5 - n)

n = 1 \sum \infty \frac{1}{2 + e ^{- n}}

What is the sum from n=2 to infinity of 4/(nln(n)) ?
The sum from n=2 to infinity of 4/(nln(n)) is diverges