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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply Series Integral Test:

diverges

= - diverges

= diverges

n = 3 \sum \infty \frac{7}{n} - \frac{7}{n + 2}

k = 0 \sum \infty \frac{k}{4 k + 1}

n = 0 \sum \infty \frac{n}{8 n ^{6} + 1}

k = 1 \sum \infty 9 (\frac{2}{3})^{k}

n = 1 \sum \infty 6 (\frac{1}{5})^{n - 1}

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