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

k = 2 \sum \infty \frac{k}{ln ( 5 k )}

n = 2 \sum \infty (\frac{1}{9 n - 2})^{7}

x = 0 \sum \infty x (x)

n = 1 \sum \infty cos (\frac{20}{n})

n = 1 \sum \infty 5 - 6 (0.7)^{n}

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