What is the sum from k=3 to infinity of 9/(kln^2(k)) ?

The sum from k=3 to infinity of 9/(kln^2(k)) is converges

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sum from k=3 to infinity of 9/(kln^2(k))

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Integral Test:

converges

= 9 converges

= converges

n = 0 \sum \infty 1 e^{- 5 n}

n = 0 \sum \infty (\frac{8}{5})^{n}

k = 2 \sum \infty (- 0.5)^{k}

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

n = 0 \sum \infty n^{4}

What is the sum from k=3 to infinity of 9/(kln^2(k)) ?
The sum from k=3 to infinity of 9/(kln^2(k)) is converges