What is the sum from k=1 to infinity of 2/(kln(9k)) ?

The sum from k=1 to infinity of 2/(kln(9k)) is diverges

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sum from k=1 to infinity of 2/(kln(9k))

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

= 2 \cdot k = 1 \sum \infty \frac{1}{k ln ( 9 k )}

Apply Series Integral Test:

diverges

= 2 diverges

= diverges

n = 8 \sum \infty \frac{1}{n} - \frac{1}{n + 3}

n = 3 \sum \infty (1 + n)^{- 1}

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

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

n = 0 \sum \infty \frac{1}{( n + 2 ) n !}

What is the sum from k=1 to infinity of 2/(kln(9k)) ?
The sum from k=1 to infinity of 2/(kln(9k)) is diverges