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

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

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

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

converges

Solution steps

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

Apply Series Integral Test:

converges

= converges

i = 1 \sum \infty \frac{1}{3 ^{i}}

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

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

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

n = 3 \sum \infty \frac{1}{n ( ln ( n ) )}

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