What is the sum from n=1 to infinity of 5/(nln(3n)) ?

The sum from n=1 to infinity of 5/(nln(3n)) is diverges

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sum from n=1 to infinity of 5/(nln(3n))

n = 1 \sum \infty \frac{5}{n ln ( 3 n )}

diverges

Solution steps

n = 1 \sum \infty \frac{5}{n ln ( 3 n )}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{1}{n ln ( 3 n )}

Apply Series Integral Test:

diverges

= 5 diverges

= diverges

n = 1 \sum \infty \frac{9}{n !}

n = 0 \sum \infty \frac{n}{5 ^{n}}

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

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

n = 1 \sum \infty \frac{1}{8 + e ^{- n}}

What is the sum from n=1 to infinity of 5/(nln(3n)) ?
The sum from n=1 to infinity of 5/(nln(3n)) is diverges