What is the sum from k=4 to infinity of 3/(kln^2(k)) ?

The sum from k=4 to infinity of 3/(kln^2(k)) is converges

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sum from k=4 to infinity of 3/(kln^2(k))

k = 4 \sum \infty \frac{3}{k ln ^{2} ( k )}

converges

Solution steps

k = 4 \sum \infty \frac{3}{k ln ^{2} ( k )}

Apply the constant multiplication rule:

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

= 3 \cdot k = 4 \sum \infty \frac{1}{k ln ^{2} ( k )}

Apply Series Integral Test:

converges

= 3 converges

= converges

n = 1 \sum \infty \frac{5 n}{n ^{2} + 1}

n = 1 \sum \infty 8 - 6 (0.8)^{n}

n = 0 \sum \infty \frac{1}{2 n ^{2} + n}

n = 2 \sum \infty n e^{- 5 n^{2}}

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

What is the sum from k=4 to infinity of 3/(kln^2(k)) ?
The sum from k=4 to infinity of 3/(kln^2(k)) is converges