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

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

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

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

diverges

Solution steps

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

Simplify

\frac{1}{ln ( n )} n : \frac{n}{ln ( n )}

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

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty - 5

n = 0 \sum \infty \frac{n}{n ^{4} + 9}

n = 1 \sum \infty \frac{1}{6 n - 1}

n = 1 \sum \infty \frac{n ^{4}}{6 ^{n}}

k = 0 \sum \infty (- \frac{1}{9})^{k}

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