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

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

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

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

diverges

Solution steps

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

Simplify

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

Apply Series Integral Test:

diverges

= diverges

n = 2 \sum \infty \frac{3}{n ( n + 3 )}

n = 1 \sum \infty \frac{1}{n ^{2} + 30}

n = 0 \sum \infty \frac{1}{( 1 - n ) ^{2}}

k = 1 \sum \infty \frac{ln ( k )}{k}

n = 1 \sum \infty - 9 (\frac{1}{6})^{n - 1}

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