What is the sum from k=5 to infinity of 6/(kln^2(k)) ?

The sum from k=5 to infinity of 6/(kln^2(k)) is converges

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sum from k=5 to infinity of 6/(kln^2(k))

k = 5 \sum \infty \frac{6}{k ln ^{2} ( k )}

converges

Solution steps

k = 5 \sum \infty \frac{6}{k ln ^{2} ( k )}

Apply the constant multiplication rule:

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

= 6 \cdot k = 5 \sum \infty \frac{1}{k ln ^{2} ( k )}

Apply Series Integral Test:

converges

= 6 converges

= converges

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

n = 0 \sum \infty \frac{n}{11 n ^{6} + 1}

n = 0 \sum \infty (0.25)^{n}

n = 0 \sum \infty \frac{n}{7 ^{n}}

n = 5 \sum \infty \frac{1}{( n - 2 ) ^{4}}

What is the sum from k=5 to infinity of 6/(kln^2(k)) ?
The sum from k=5 to infinity of 6/(kln^2(k)) is converges