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

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

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

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

n = 1 \sum \infty \frac{1}{n ^{2} + 16}

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

n = 0 \sum \infty (\frac{5}{8})^{n}

n = 0 \sum \infty \frac{e ^{n}}{n ^{3}}

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

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