What is the sum from k=3 to infinity of 7/(kln^2(k)) ?

The sum from k=3 to infinity of 7/(kln^2(k)) is converges

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sum from k=3 to infinity of 7/(kln^2(k))

k = 3 \sum \infty \frac{7}{k ln ^{2} ( k )}

converges

Solution steps

k = 3 \sum \infty \frac{7}{k ln ^{2} ( k )}

Apply the constant multiplication rule:

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

= 7 \cdot k = 3 \sum \infty \frac{1}{k ln ^{2} ( k )}

Apply Series Integral Test:

converges

= 7 converges

= converges

What is the sum from k=3 to infinity of 7/(kln^2(k)) ?
The sum from k=3 to infinity of 7/(kln^2(k)) is converges