What is the sum from k=3 to infinity of 1/(ln(k)) ?

The sum from k=3 to infinity of 1/(ln(k)) is diverges

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sum from k=3 to infinity of 1/(ln(k))

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

\int \frac{1}{x 1 - 4 ln ^{2} ( x )} d x

f (x) = \frac{20 x}{x ^{2} - 5}, at x = 5

\int \frac{1}{2} sec^{2} (x) + 4 x d x

\frac{d ^{3}}{d x ^{3}} (2 x + 8)

\int_{- 5}^{0} (25 - x^{2}) d x

What is the sum from k=3 to infinity of 1/(ln(k)) ?
The sum from k=3 to infinity of 1/(ln(k)) is diverges