What is the sum from n=3 to infinity of 4/(ln(n)) ?

The sum from n=3 to infinity of 4/(ln(n)) is diverges

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sum from n=3 to infinity of 4/(ln(n))

n = 3 \sum \infty \frac{4}{ln ( n )}

diverges

Solution steps

n = 3 \sum \infty \frac{4}{ln ( n )}

Apply the constant multiplication rule:

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

= 4 \cdot n = 3 \sum \infty \frac{1}{ln ( n )}

Apply Series Integral Test:

diverges

= 4 diverges

= diverges

n = 0 \sum \infty \frac{1}{7}

n = 0 \sum \infty \frac{1}{8}

n = 4 \sum \infty \frac{1}{3 ^{n - 3}}

n = 1 \sum \infty n! e^{- 9 n}

n = 1 \sum \infty (3.5 + 1)^{n}

What is the sum from n=3 to infinity of 4/(ln(n)) ?
The sum from n=3 to infinity of 4/(ln(n)) is diverges