What is the sum from n=2 to infinity of 5/(nln(n)) ?

The sum from n=2 to infinity of 5/(nln(n)) is diverges

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sum from n=2 to infinity of 5/(nln(n))

n = 2 \sum \infty \frac{5}{n ln ( n )}

diverges

Solution steps

n = 2 \sum \infty \frac{5}{n ln ( n )}

Apply the constant multiplication rule:

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

= 5 \cdot n = 2 \sum \infty \frac{1}{n ln ( n )}

Apply Series Integral Test:

diverges

= 5 diverges

= diverges

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

n = 1 \sum \infty \frac{1}{5 n - 1}

n = 2 \sum \infty (- 0.6)^{n}

n = 1 \sum \infty cos (\frac{14}{n})

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

What is the sum from n=2 to infinity of 5/(nln(n)) ?
The sum from n=2 to infinity of 5/(nln(n)) is diverges