What is the sum from n=2 to infinity of 7/(nln(n)) ?

The sum from n=2 to infinity of 7/(nln(n)) is diverges

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sum from n=2 to infinity of 7/(nln(n))

n = 2 \sum \infty \frac{7}{n ln ( n )}

diverges

Solution steps

n = 2 \sum \infty \frac{7}{n ln ( n )}

Apply the constant multiplication rule:

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

= 7 \cdot n = 2 \sum \infty \frac{1}{n ln ( n )}

Apply Series Integral Test:

diverges

= 7 diverges

= diverges

n = 0 \sum \infty \frac{n !}{40 0 ^{n}}

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

n = 1 \sum \infty cos (\frac{6}{n})

n = 1 \sum \infty n^{\frac{1}{2}}

n = 0 \sum \infty (\frac{1}{15})^{n}

What is the sum from n=2 to infinity of 7/(nln(n)) ?
The sum from n=2 to infinity of 7/(nln(n)) is diverges