What is the sum from n=5 to infinity of 1/(n^2ln(n)) ?

The sum from n=5 to infinity of 1/(n^2ln(n)) is converges

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sum from n=5 to infinity of 1/(n^2ln(n))

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

converges

Solution steps

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

Apply Series Integral Test:

converges

= converges

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

n = 1 \sum \infty \frac{7}{n ^{\frac{2}{5}}}

n = 1 \sum \infty \frac{6 n}{3 n + 9}

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

n = 0 \sum \infty cos (\frac{nπ}{3})

What is the sum from n=5 to infinity of 1/(n^2ln(n)) ?
The sum from n=5 to infinity of 1/(n^2ln(n)) is converges