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

The sum from n=5 to infinity of 1/(ln(n)) is diverges

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

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

n = 0 \sum \infty n^{- 1}

n = 0 \sum \infty 2 - 3 (0.8)^{n}

n = 1 \sum \infty \frac{6}{n ^{2} + n ^{3}}

n = 1 \sum \infty \frac{9}{n ^{2} + 4}

n = 1 \sum \infty \frac{n ^{5}}{8 ^{n}}

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