What is the sum from n=2 to infinity of (10)/(ln(n)) ?

The sum from n=2 to infinity of (10)/(ln(n)) is diverges

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sum from n=2 to infinity of (10)/(ln(n))

n = 2 \sum \infty \frac{10}{ln ( n )}

diverges

Solution steps

n = 2 \sum \infty \frac{10}{ln ( n )}

Apply the constant multiplication rule:

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

= 10 \cdot n = 2 \sum \infty \frac{1}{ln ( n )}

Apply Series Integral Test:

diverges

= 10 diverges

= diverges

n = 0 \sum \infty \frac{9}{2} (\frac{1}{2})^{n}

n = 0 \sum \infty \frac{n ^{2} + 1}{n}

k = 0 \sum \infty (\frac{1}{2})^{k}

n = 0 \sum \infty \frac{10 n}{4 n + 1}

n = 1 \sum \infty \frac{9}{5 n ^{3}}

What is the sum from n=2 to infinity of (10)/(ln(n)) ?
The sum from n=2 to infinity of (10)/(ln(n)) is diverges